ICSE Class 10 Trigonometrical Identities

Q: What are the three fundamental trigonometrical identities?

The three fundamental Pythagorean identities are: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ. Most proofs in this chapter rely on manipulating expressions to use one of these core identities.

Q: How do you prove a trigonometrical identity?

To prove an identity, you typically start with the more complex side of the equation, usually the Left-Hand Side (LHS). You then use algebraic manipulation and fundamental identities to simplify it step-by-step until it becomes identical to the other side, the Right-Hand Side (RHS).

Q: What is the difference between a trigonometric equation and an identity?

A trigonometric identity is true for all possible values of the variable angle, whereas a trigonometric equation is only true for specific values of the variable. For example, sin²θ + cos²θ = 1 is an identity, but sin θ = 1/2 is an equation.

Q: Are these solutions enough for the ICSE board exams?

Yes, these solutions are prepared by subject matter experts and follow the exact step-by-step method required by the ICSE board. Practising these problems will prepare you thoroughly for the types of questions on trigonometrical identities asked in board examinations.

This chapter provides the complete ICSE Class 10 Trigonometrical Identities Solutions from the popular textbook, Class – 10 Concise Mathematics Selina. Mastering trigonometrical identities is a key step in your Maths journey, as it involves proving that a given equation involving trigonometric functions holds true for all values of the angle. You will learn to use the three fundamental Pythagorean identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ. The exercises here will train you to manipulate complex expressions by converting them into sines and cosines, rationalising, or using algebraic formulas to simplify and prove the required results. These skills are absolutely essential for higher-level mathematics.

If you are stuck on a proof or want to check your method for solving a problem, you have landed on the right page. We provide detailed, step-by-step solutions for all 150 questions found in Exercise 21(A), Exercise 21(B), Exercise 21(C), Exercise 21(D), and the final Test Yourself section. Each solution is carefully crafted to follow the exact methodology and presentation style that the ICSE board expects in your exams. Here, you will find clear, accurate, and easy-to-follow proofs to help you master every question in the Trigonometrical Identities chapter.

Exercise 21(A)

Question 1(a)

sin^4 θ – cos^4 θ is equal to :

(a) sin^2 θ – 1
(b) 1 + cos^2 θ
(c) 2 sin^2 θ – 1
(d) 1 – cos^2 θ

Answer: (c) 2 sin^2 θ – 1

Consider the expression $\sin^4 \theta - \cos^4 \theta$ . Notice that this can be rewritten using the difference of squares formula:

⇒ $(\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta)$

Since we know $\sin^2 \theta + \cos^2 \theta = 1$ , substitute this value into the equation:

⇒ $1 \times (\sin^2 \theta - \cos^2 \theta)$

This simplifies to:

⇒ $\sin^2 \theta - \cos^2 \theta$

Now, replace $\cos^2 \theta$ with $1 - \sin^2 \theta$ :

⇒ $\sin^2 \theta - (1 - \sin^2 \theta)$

Simplifying further gives:

⇒ $\sin^2 \theta + \sin^2 \theta - 1$

⇒ $2 \sin^2 \theta - 1$ .

Hence, Option 3 is the correct option.

Question 1(b)

(1 + tan θ)^2 + (1 – tan θ)^2 is equal to :

(a) 2 cosec^2 θ
(b) 2 sec^2 θ
(c) cosec^2 θ
(d) sec^2 θ

Answer: (b) 2 sec^2 θ

Consider the expression:

⇒ $(1 + \tan \theta)^2 + (1 - \tan \theta)^2$

Expanding each square, we have:

⇒ $1 + \tan^2 \theta + 2 \tan \theta + 1 + \tan^2 \theta - 2 \tan \theta$

Notice that $+2 \tan \theta$ and $-2 \tan \theta$ cancel each other out, leaving us with:

⇒ $2 + 2 \tan^2 \theta$

Factor out the common factor of 2:

⇒ $2(1 + \tan^2 \theta)$

Recall the identity $1 + \tan^2 \theta = \sec^2 \theta$ . Substituting this, we get:

⇒ $2 \sec^2 \theta$ .

Hence, Option 2 is the correct option.

Question 1(c)

cot^4 θ + cot^2 θ is equal to :

(a) 2 cot^2 θ. cosec^2 θ
(b) tan^2 θ + tan^4 θ
(c) tan^2 θ.cosec^2 θ
(d) cosec^4 θ – cosec^2 θ

Answer: (d) cosec^4 θ – cosec^2 θ

Consider the expression $\cot^4 \theta + \cot^2 \theta$ .

First, factor out $\cot^2 \theta$ :

\cot^4 \theta + \cot^2 \theta = \cot^2 \theta (\cot^2 \theta + 1)

Now, recall the identity $\cot^2 \theta = \cosec^2 \theta - 1$ . Substitute this into the expression:

\cot^2 \theta (\cot^2 \theta + 1) = (\cosec^2 \theta - 1) \cdot \cosec^2 \theta

Distributing $\cosec^2 \theta$ gives:

\cosec^4 \theta - \cosec^2 \theta

∴ The expression simplifies to $\cosec^4 \theta - \cosec^2 \theta$ .

Hence, Option 4 is the correct option.

Question 1(d)

$\dfrac{1}{\text{1 - sin A}}$ is equal to :

(a) 1 + sin A
(b) sec^2 A (1 + sin A)
(c) sec^2 A
(d) sec^2 A + sin A

Answer: (b) sec^2 A (1 + sin A)

Consider the expression:

⇒ $\dfrac{1}{\text{1 - sin A}}$

To simplify, multiply both the numerator and the denominator by $(1 + \text{sin A})$ :

\Rightarrow \dfrac{1}{\text{1 - sin A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}}

This results in:

\Rightarrow \dfrac{\text{1 + sin A}}{1 - \text{sin}^2 A}

Recall that $1 - \text{sin}^2 A = \text{cos}^2 A$ by the Pythagorean identity. Thus, we have:

\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos}^2 A}

Breaking this down further, it can be expressed as:

\Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{\text{sin A}}{\text{cos}^2 A}

Which simplifies to:

\Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{cos}^2 A} \cdot \text{sin A}

This can be rewritten using the secant function as:

\Rightarrow \text{sec}^2 A + \text{sec}^2 A \cdot \text{sin A}

Factoring out $\text{sec}^2 A$ , we get:

\Rightarrow \text{sec}^2 A (1 + \text{sin A})

Thus, the expression simplifies to $\text{sec}^2 A (1 + \text{sin A})$ .

Hence, Option 2 is the correct option.

Question 1(e)

$\dfrac{\text{tan}^2 A}{\text{(sec A + 1)}^2}$ is equal to :

(a) $\dfrac{\text{1 + cos A}}{\text{1 - cos A}}$
(b) $\dfrac{\text{1 - cos A}}{\text{1 + cos A}}$
(c) $\dfrac{1}{\text{1 + cos A}}$
(d) $\dfrac{1}{\text{1 - cos A}}$

Answer: (b)

\dfrac{\text{1 - cos A}}{\text{1 + cos A}}

We start with the expression $\dfrac{\text{tan}^2 A}{(\text{sec A} + 1)^2}$ . Notice that $\text{tan}^2 A$ can be rewritten using the identity $\text{tan}^2 A = \text{sec}^2 A - 1$ . Thus, the expression becomes:

\Rightarrow \dfrac{\text{sec}^2 A - 1}{(\text{sec A} + 1)^2}

Recognize that $\text{sec}^2 A - 1$ can be factored as $(\text{sec A} + 1)(\text{sec A} - 1)$ . So, we have:

\Rightarrow \dfrac{(\text{sec A} + 1)(\text{sec A} - 1)}{(\text{sec A} + 1)^2}

Cancel out the common factor $\text{(sec A + 1)}$ in the numerator and denominator:

\Rightarrow \dfrac{\text{sec A} - 1}{\text{sec A} + 1}

Now, express $\text{sec A}$ in terms of $\text{cos A}$ , where $\text{sec A} = \dfrac{1}{\text{cos A}}$ :

\Rightarrow \dfrac{\dfrac{1}{\text{cos A}} - 1}{\dfrac{1}{\text{cos A}} + 1}

Simplify the fractions in the numerator and denominator:

\Rightarrow \dfrac{\dfrac{1 - \text{cos A}}{\text{cos A}}}{\dfrac{1 + \text{cos A}}{\text{cos A}}}

The $\text{cos A}$ terms cancel out, leaving:

\Rightarrow \dfrac{1 - \text{cos A}}{1 + \text{cos A}}

Hence, option 2 is the correct option.

Question 2

Prove the following identities :

\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}

Answer:

To establish the identity, let’s start with the left-hand side (L.H.S.) of the equation:

\Rightarrow \dfrac{\text{sec A - 1}}{\text{sec A + 1}}

Recall that $\text{sec A} = \dfrac{1}{\text{cos A}}$ . Substituting this, we have:

\Rightarrow \dfrac{\dfrac{1}{\text{cos A}} - 1}{\dfrac{1}{\text{cos A}} + 1}

Simplifying further by finding a common denominator for the numerator and denominator:

\Rightarrow \dfrac{\dfrac{\text{1 - cos A}}{\text{cos A}}}{\dfrac{\text{1 + cos A}}{\text{cos A}}}

Now, multiply the numerator and the denominator by $\text{cos A}$ :

\Rightarrow \dfrac{\text{(1 - cos A) × cos A}}{\text{(1 + cos A) × cos A}}

The $\text{cos A}$ terms cancel out, resulting in:

\Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}.

Thus, we have shown that L.H.S. equals the right-hand side (R.H.S.).

Hence, proved that $\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}$ .

Question 3

Prove the following identities :

\dfrac{1}{\text{tan A + cot A}} = \text{cos A sin A}

Answer:

Let’s work through the left-hand side (L.H.S.) of the given identity:

\Rightarrow \dfrac{1}{\text{tan A + cot A}}

Now, express tan A and cot A in terms of sine and cosine:

\Rightarrow \dfrac{1}{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}}

Combine the fractions in the denominator:

\Rightarrow \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}}

Taking the reciprocal:

\Rightarrow \dfrac{\text{sin A cos A}}{\text{sin}^2 A + \text{cos}^2 A}

Recall the Pythagorean identity:

\text{sin}^2 A + \text{cos}^2 A = 1

Substitute this into the expression:

\Rightarrow \text{sin A cos A}

Thus, the L.H.S. simplifies to the right-hand side (R.H.S.), showing that both sides are equal.

Hence, proved that $\dfrac{1}{\text{tan A + cot A}} = \text{cos A sin A}$ .

Question 4

Prove the following identities :

tan A – cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Answer:

Let’s start with the left-hand side of the identity:

\Rightarrow \ \text{tan A - cot A}

This can be expressed using sine and cosine:

\Rightarrow \ \dfrac{\text{sin A}}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}}

Combining these fractions over a common denominator gives:

\Rightarrow \ \dfrac{\text{sin}^2 A - \text{cos}^2 A}{\text{cos A sin A}}

Recall the Pythagorean identity:

\text{sin}^2 A = 1 - \text{cos}^2 A

Substituting this into our expression, we get:

\Rightarrow \ \dfrac{1 - \text{cos}^2 A - \text{cos}^2 A}{\text{sin A cos A}}

Simplifying further, we arrive at:

\Rightarrow \ \dfrac{1 - \text{2 cos}^2 A}{\text{sin A cos A}}.

Thus, we see that the left-hand side equals the right-hand side.

Hence, proved that tan A – cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Question 5

Prove the following identities :

cosec^4 A – cosec^2 A = cot^4 A + cot^2 A

Answer:

Let’s focus on the left-hand side of the equation:

⇒ $\text{cosec}^4 A - \text{cosec}^2 A$

Factor out $\text{cosec}^2 A$ :

⇒ ( \text{cosec}^2 A(\text{cosec}^2 A – 1) )

Recall the identity:

\text{cosec}^2 A = 1 + \text{cot}^2 A

Substitute this into the expression:

⇒ ( (1 + \text{cot}^2 A)(1 + \text{cot}^2 A – 1) )

Simplify the terms inside the parentheses:

⇒ ( (1 + \text{cot}^2 A) \times \text{cot}^2 A )

Distribute $\text{cot}^2 A$ :

⇒ $\text{cot}^2 A + \text{cot}^4 A$ .

Thus, the left-hand side equals the right-hand side, confirming that:

Hence, proved that $\text{cosec}^4 A - \text{cosec}^2 A = \text{cot}^4 A + \text{cot}^2 A$ .

Question 6

Prove the following identities :

sec A(1 – sin A)(sec A + tan A) = 1

Answer:

Let’s work through the left-hand side of the equation:

\begin{aligned}\Rightarrow \text{sec A(1 - sin A)(sec A + tan A)} \\\Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big) \\\Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big) \\\Rightarrow \dfrac{1 - \text{sin}^2 A}{\text{cos}^2A}\end{aligned}

Notice how we use the identity $\text{cos}^2 A = 1 - \text{sin}^2 A$ .

This gives us:

\begin{aligned}\Rightarrow \dfrac{1 - \text{sin}^2 A}{1 - \text{sin}^2A} \\\Rightarrow 1.\end{aligned}

Thus, the left-hand side equals the right-hand side.

Hence, proved that sec A(1 – sin A)(sec A + tan A) = 1.

Question 7

Prove the following identities :

sec^2 A + cosec^2 A = sec^2 A . cosec^2 A

Answer:

To tackle the left-hand side of the identity, we start with:

\Rightarrow \text{sec}^2 A + \text{cosec}^2 A

This can be rewritten using the definitions of secant and cosecant:

\Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A}

Combine these fractions over a common denominator:

\Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A \cdot \text{sin}^2 A}

We know from the Pythagorean identity that $\text{sin}^2 A + \text{cos}^2 A = 1$ . Substitute this result:

\Rightarrow \dfrac{1}{\text{cos}^2 A \cdot \text{sin}^2 A}

This expression is precisely the product of secant squared and cosecant squared:

\Rightarrow \text{sec}^2 A \cdot \text{cosec}^2 A

Thus, the left-hand side equals the right-hand side.

Hence, proved that sec^2 A + cosec^2 A = sec^2 A . cosec^2 A.

Question 8

Prove the following identities :

\dfrac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}

Answer:

To demonstrate this identity, let’s start with the left-hand side:

\Rightarrow \dfrac{\Big(1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}}}{\dfrac{1}{\text{sin}^2 A}}

This simplifies to:

\Rightarrow \Big(\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}} \times \text{sin}^2 A

Using the Pythagorean identity, we know:

\cos^2 A + \sin^2 A = 1

Thus, the expression becomes:

\Rightarrow \dfrac{1}{\text{cos}^2 A} \times \text{cos A} \times \text{sin A}

Simplifying further, we have:

\Rightarrow \dfrac{\text{sin A}}{\text{cos A}}

Which is equivalent to:

\Rightarrow \text{tan A}.

Therefore, the left-hand side equals the right-hand side, confirming the identity:

Hence, proved that $\dfrac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}$ .

Question 9

Prove the following identities :

tan^2 A – sin^2 A = tan^2 A. sin^2 A

Answer:

We start by simplifying the left-hand side of the identity:

\Rightarrow \text{tan}^2 A - \text{sin}^2 A

Express $\text{tan}^2 A$ in terms of sine and cosine:

\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \text{sin}^2 A

Factor out $\text{sin}^2 A$ :

\Rightarrow \text{sin}^2 A \times \Big(\dfrac{1}{\text{cos}^2 A} - 1\Big)

Rearrange the terms inside the bracket:

\Rightarrow \text{sin}^2 A \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos}^2 A}\Big)

Using the identity $1 - \text{cos}^2 A = \text{sin}^2 A$ , we substitute:

\Rightarrow \text{sin}^2 A \times \dfrac{\text{sin}^2 A}{\text{cos}^2 A}

This simplifies to:

\Rightarrow \text{sin}^2 A. \text{tan}^2 A

Thus, the left-hand side equals the right-hand side.

Hence, proved that tan^2 A – sin^2 A = tan^2 A. sin^2 A.

Question 10

Prove the following identities :

(cosec A + sin A)(cosec A – sin A) = cot^2 A + cos^2 A

Answer:

To start, let’s recall the identity:

\cosec^2 A = 1 + \cot^2 A

and another identity:

\sin^2 A = 1 - \cos^2 A

Now, we will work through the left-hand side of the given equation:

(\cosec A + \sin A)(\cosec A - \sin A)

This expression can be expanded using the difference of squares formula:

\Rightarrow \cosec^2 A - \sin^2 A

Substituting the identities, we have:

\Rightarrow (1 + \cot^2 A) - (1 - \cos^2 A)

Simplifying further, we get:

\Rightarrow 1 + \cot^2 A - 1 + \cos^2 A

which simplifies to:

\Rightarrow \cot^2 A + \cos^2 A

Thus, it is proven that ((\cosec A + \sin A)(\cosec A – \sin A) = \cot^2 A + \cos^2 A).

Question 11

Prove the following identities :

(cos A + sin A)^2 + (cos A – sin A)^2 = 2

Answer:

Consider the left-hand side of the identity:

⇒ $(\cos A + \sin A)^2 + (\cos A - \sin A)^2$

Expanding both squares, we have:

⇒ $\cos^2 A + \sin^2 A + 2 \cos A \sin A + \cos^2 A + \sin^2 A - 2 \cos A \sin A$

Notice that $2 \cos A \sin A$ and $-2 \cos A \sin A$ cancel each other out, leaving:

⇒ $2(\sin^2 A + \cos^2 A)$

We know from the Pythagorean identity that $\sin^2 A + \cos^2 A = 1$ .

Substituting this, we get:

⇒ $2 \times 1$

⇒ $2$ .

Thus, the left-hand side equals the right-hand side.

Hence, proved that $(\cos A + \sin A)^2 + (\cos A - \sin A)^2 = 2$ .

Question 12

Prove the following identities :

(cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1

Answer:

To tackle the left-hand side of the identity, start by expanding each term:

\Rightarrow (\text{cosec A} - \text{sin A})(\text{sec A} - \text{cos A})(\text{tan A} + \text{cot A})

Replace each trigonometric function with its reciprocal form:

\Rightarrow \left(\dfrac{1}{\text{sin A}} - \text{sin A}\right) \times \left(\dfrac{1}{\text{cos A}} - \text{cos A}\right) \times \left(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\right)

Simplify using the Pythagorean identities:

\Rightarrow \left(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\right) \times \left(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\right) \times \left(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A} \cdot \text{sin A}}\right)

Recall the identities:
– $1 - \text{sin}^2 A = \text{cos}^2 A$
– $1 - \text{cos}^2 A = \text{sin}^2 A$
– $\text{sin}^2 A + \text{cos}^2 A = 1$

Substitute these into the expression:

\Rightarrow \left(\dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos A} \cdot \text{sin A}}\right)

Simplify the expression:

\Rightarrow \dfrac{\text{cos}^2 A \cdot \text{sin}^2 A}{\text{cos}^2 A \cdot \text{sin}^2 A}

This simplifies to:

\Rightarrow 1.

∴ The left-hand side is equal to the right-hand side.

Hence, proved that (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1.

Question 13

Prove the following identities :

$\dfrac{1}{\text{sec A + tan A}}$ = sec A – tan A

Answer:

Consider the left-hand side of the identity:

\Rightarrow \dfrac{1}{\text{sec A + tan A}}

To simplify, multiply both the numerator and the denominator by $(\text{sec A - tan A})$ :

\begin{aligned}\Rightarrow \dfrac{1}{\text{sec A + tan A}} \times \dfrac{\text{sec A - tan A}}{\text{sec A - tan A}} \\\Rightarrow \dfrac{\text{sec A - tan A}}{\text{sec}^2 A - \text{tan}^2 A}\end{aligned}

Recall the identity:

\text{sec}^2 A - \text{tan}^2 A = 1

Substitute this into the equation:

\begin{aligned}\Rightarrow \dfrac{\text{sec A - tan A}}{1} \\\Rightarrow \text{sec A - tan A}.\end{aligned}

Thus, we have shown that the left-hand side equals the right-hand side.

Hence proved that $\dfrac{1}{\text{sec A + tan A}}$ = sec A – tan A.

Question 14

Prove the following identities :

(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A

Answer:

We start with some fundamental trigonometric identities:

$\sin^2 A + \cos^2 A = 1$
$\sec^2 A = 1 + \tan^2 A$
$\cosec^2 A = 1 + \cot^2 A$

Now, let’s work through the left-hand side (L.H.S.) of the equation:

(\sin A + \cosec A)^2 + (\cos A + \sec A)^2

Expanding both squares, we have:

\sin^2 A + \cosec^2 A + 2 \sin A \cdot \cosec A + \cos^2 A + \sec^2 A + 2 \cos A \cdot \sec A

Substitute the identities and simplify:

\sin^2 A + 1 + \cot^2 A + 2 \times \sin A \times \dfrac{1}{\text{sin A}} + \cos^2 A + 1 + \tan^2 A + 2 \times \cos A \times \dfrac{1}{\text{cos A}}

This simplifies to:

\sin^2 A + \cos^2 A + 1 + \cot^2 A + 2 + 1 + \tan^2 A + 2

Since $\sin^2 A + \cos^2 A = 1$ , it becomes:

1 + 1 + 2 + 1 + 2 + \cot^2 A + \tan^2 A

Finally, we obtain:

7 + \tan^2 A + \cot^2 A

Thus, the left-hand side equals the right-hand side (R.H.S.).

Hence, proved that $(\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ .

Question 15

Prove the following identities :

sec^2 A . cosec^2 A = tan^2 A + cot^2 A + 2

Answer:

Let’s consider the left-hand side of the identity:

⇒ $\text{sec}^2 A . \text{cosec}^2 A$

⇒ $\dfrac{1}{\text{cos}^2 A . \text{sin}^2 A}$ .

Now, let’s work on the right-hand side:

\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A} + 2

This can be expressed as:

\Rightarrow \dfrac{\text{sin}^4 A + \text{cos}^4 A + 2 \text{sin}^2 A \text{cos}^2 A}{\text{cos}^2 A \text{sin}^2 A}

Notice that:

\Rightarrow \dfrac{(\text{sin}^2 A + \text{cos}^2 A)^2}{\text{cos}^2 A \text{sin}^2 A}

Using the fundamental identity $\text{sin}^2 A + \text{cos}^2 A = 1$ , we have:

\Rightarrow \dfrac{(1)^2}{\text{cos}^2 A \text{sin}^2 A}

\Rightarrow \dfrac{1}{\text{cos}^2 A \text{sin}^2 A}.

Thus, we observe that both the left-hand side and the right-hand side simplify to $\dfrac{1}{\text{cos}^2 A \text{sin}^2 A}$ .

Hence, proved that $\text{sec}^2 A . \text{cosec}^2 A = \text{tan}^2 A + \text{cot}^2 A + 2$ .

Question 16

Prove the following identities :

$\dfrac{1}{\text{1 + cos A}} + \dfrac{1}{\text{1 - cos A}}$ = 2 cosec^2 A

Answer:

Consider the left-hand side of the equation:

\Rightarrow \dfrac{1}{1 + \cos A} + \dfrac{1}{1 - \cos A}

To simplify, find a common denominator:

\Rightarrow \dfrac{(1 - \cos A) + (1 + \cos A)}{(1 + \cos A)(1 - \cos A)}

Notice that the numerator simplifies to:

\Rightarrow \dfrac{2}{1 - \cos^2 A}

Recall the Pythagorean identity:

1 - \cos^2 A = \sin^2 A

Substituting this into the expression gives:

\Rightarrow \dfrac{2}{\sin^2 A}

Recognizing that $\dfrac{1}{\sin^2 A}$ is $\text{cosec}^2 A$ , we conclude:

\Rightarrow 2\text{ cosec}^2 A.

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{1}{1 + \cos A} + \dfrac{1}{1 - \cos A}$ = 2 cosec^2 A.

Question 17

Prove the following identities :

$\dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}}$ = 2 sec^2 A

Answer:

Let’s examine the left-hand side of the identity:

\Rightarrow \dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}}

To simplify, we find a common denominator:

\Rightarrow \dfrac{\text{cosec A(cosec A + 1) + cosec A(cosec A - 1)}}{\text{(cosec A - 1)(cosec A + 1)}}

Expanding the terms in the numerator:

\Rightarrow \dfrac{\text{cosec}^2 A + \text{cosec A + cosec}^2 A - \text{cosec A}}{\text{cosec}^2 A - 1}

Simplify by combining like terms:

\Rightarrow \dfrac{\text{2 cosec}^2 A}{\text{cot}^2 A}

Notice that $\text{cosec}^2 A = \dfrac{1}{\text{sin}^2 A}$ and $\text{cot}^2 A = \dfrac{\text{cos}^2 A}{\text{sin}^2 A}$ :

\Rightarrow \dfrac{2 \times \dfrac{1}{\text{sin}^2 A}}{\dfrac{\text{cos}^2 A}{\text{sin}^2 A}}

This simplifies to:

\Rightarrow \dfrac{2 \times \dfrac{1}{\text{sin}^2 A} \times \text{sin}^2 A}{\text{cos}^2 A}

Cancel out $\text{sin}^2 A$ :

\Rightarrow \dfrac{2}{\text{cos}^2 A}

Recognizing that $\dfrac{1}{\text{cos}^2 A} = \text{sec}^2 A$ , we have:

\Rightarrow 2\text{sec}^2 A.

Thus, we have shown that the left-hand side equals the right-hand side. Hence, proved that $\dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}}$ = 2 sec^2 A.

Question 18

Prove the following identities :

\dfrac{\text{1 + cos A}}{\text{1 - cos A}} = \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2}

Answer:

To demonstrate the identity, let’s focus on the right-hand side:

\Rightarrow \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2}

We know that $\text{tan}^2 A = \dfrac{\text{sin}^2 A}{\text{cos}^2 A}$ . Substituting this, we have:

\Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\dfrac{1}{\text{cos A}} - 1\Big)^2}

Simplify the expression in the denominator:

\Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\dfrac{\text{1 - cos A}}{\text{cos} A}\Big)^2}

Now, multiply the numerator by the reciprocal of the denominator:

\Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 - cos A)}^2}

This simplifies to:

\Rightarrow \dfrac{\text{sin}^2 A}{\text{(1 - cos A)}^2}

Recall that $\text{sin}^2 A = 1 - \text{cos}^2 A$ . Substitute this into the equation:

\Rightarrow \dfrac{\text{1 - cos}^2 A}{(1 - \text{ cos A})^2}

Factor the numerator:

\Rightarrow \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{(1 - cos A)}^2}

Cancel ( \text{(1 – cos A)} ) from the numerator and denominator:

\Rightarrow \dfrac{\text{(1 + cos A)}}{\text{(1 - cos A)}}

Thus, both sides of the equation match.

Hence, proved that ( \dfrac{\text{1 + cos A}}{\text{1 – cos A}} = \dfrac{\text{tan}^2 A}{\text{(sec A – 1)}^2} ).

Question 19

Prove the following identities :

$\dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}}$ = 2 sec A

Answer:

Let’s work through the left-hand side of the given identity:

\begin{aligned}\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} \\\Rightarrow \dfrac{\text{(1 + sin A)}^2 + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\\Rightarrow \dfrac{\text{1 + 2 sin A + sin}^2 A + \text{cos}^2 A}{\text{cos A(1 + sin A)}}\end{aligned}

Notice that we can use the Pythagorean identity:

$\sin^2 A + \cos^2 A = 1$ .

Substituting this into our expression, we have:

\begin{aligned}\Rightarrow \dfrac{\text{1 + 2 sin A + 1}}{\text{cos A(1 + sin A)}} \\\Rightarrow \dfrac{\text{2 + 2 sin A}}{\text{cos A(1 + sin A)}} \\\Rightarrow \dfrac{\text{2(1 + sin A)}}{\text{cos A(1 + sin A)}} \\\Rightarrow \dfrac{2}{\text{cos A}} \\\Rightarrow 2\text{ sec A}\end{aligned}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}}$ = 2 sec A.

Question 20

Prove the following identities :

$\dfrac{\text{1 - sin A}}{\text{1 + sin A}}$ = (sec A – tan A)^2

Answer:

Let’s tackle the right-hand side (R.H.S.) of the equation:

\Rightarrow \text{(sec A - tan A)}^2

This can be expressed as:

\Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2

Simplifying inside the brackets gives:

\Rightarrow \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2

This further simplifies to:

\Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{cos}^2 A}

Recall the identity:

\text{cos}^2 A = 1 - \text{sin}^2 A

Substituting this identity, we have:

\Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{1 - sin}^2 A}

This can be factored as:

\Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{(1 - sin A)(1 + sin A)}}

Upon canceling out the common term $(1 - \text{sin A})$ , we get:

\Rightarrow \dfrac{\text{1 - sin A}}{\text{1 + sin A}}.

Thus, the left-hand side (L.H.S.) equals the right-hand side (R.H.S.).

Hence, proved that $\dfrac{\text{1 - sin A}}{\text{1 + sin A}}$ = (sec A – tan A)^2.

Question 21

Prove the following identities :

\dfrac{\text{cosec A - 1}}{\text{cosec A + 1}} = \Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2

Answer:

Let’s start by examining the left-hand side (LHS) of the identity:

\dfrac{\dfrac{1}{\text{sin A}} - 1}{\dfrac{1}{\text{sin A}} + 1}

This can be rewritten as:

\dfrac{\dfrac{\text{1 - sin A}}{\text{sin A}}}{\dfrac{\text{1 + sin A}}{\text{sin A}}}

When we simplify this, we get:

\dfrac{\text{(1 - sin A)} \times \text{sin A}}{\text{(1 + sin A)} \times \text{sin A}}

Further simplification leads to:

\dfrac{\text{1 - sin A}}{\text{1 + sin A}}

Now, let’s analyze the right-hand side (RHS) of the identity:

\Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2

This expression can be expanded to:

\dfrac{\text{cos}^2 A}{\text{(1 + sin A)}^2}

Using the Pythagorean identity $\cos^2 A = 1 - \sin^2 A$ , substitute to get:

\dfrac{\text{1 - sin}^2 A}{\text{(1 + sin A)}^2}

This simplifies to:

\dfrac{\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)}^2}

Cancel out the common $1 + \sin A$ terms:

\dfrac{\text{(1 - sin A)}}{\text{(1 + sin A)}}

Notice that the LHS and RHS are identical.

Hence, proved that (\dfrac{\text{cosec A – 1}}{\text{cosec A + 1}} = \Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2).

Question 22

Prove the following identities :

tan^2 A – tan^2 B = $\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}$

Answer:

Let’s verify the identity by working on the Left Hand Side (L.H.S.) of the equation:

Start with:

\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{sin}^2 B}{\text{cos}^2 B}

Combine the fractions by finding a common denominator:

\Rightarrow \dfrac{\text{sin}^2 A \cdot \text{cos}^2 B - \text{sin}^2 B \cdot \text{cos}^2 A}{\text{cos}^2 A \cdot \text{cos}^2 B}

Notice that $1 - \text{sin}^2 B = \text{cos}^2 B$ and $1 - \text{sin}^2 A = \text{cos}^2 A$ . Use these identities:

\Rightarrow \dfrac{\text{sin}^2 A (1 - \text{sin}^2 B) - \text{sin}^2 B (1 - \text{sin}^2 A)}{\text{cos}^2 A \cdot \text{cos}^2 B}

Simplify the expression:

\Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 A \cdot \text{sin}^2 B - \text{sin}^2 B + \text{sin}^2 A \cdot \text{sin}^2 B}{\text{cos}^2 A \cdot \text{cos}^2 B}

Combine like terms:

\Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A \cdot \text{cos}^2 B}

∴ The Left Hand Side equals the Right Hand Side.

Hence, proved that tan^2 A – tan^2 B = $\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}$ .

Question 23

Prove the following identities :

$\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}}$ = tan θ

Answer:

Let’s tackle the left-hand side of the identity:

\Rightarrow \dfrac{\text{sin } \theta (1 - 2 \text{sin}^2 \theta)}{\text{cos } \theta (2 \text{cos}^2 \theta - 1)}

Using the Pythagorean identity, we know:

\text{sin}^2 \theta = 1 - \text{cos}^2 \theta

Substituting this into our expression, we have:

$\Rightarrow \dfrac{\text{sin } \theta$ 1 – 2(1 – \text{cos}^2 \theta) $}{\text{cos } \theta (2 \text{cos}^2 \theta - 1)}$

Simplifying the numerator:

\Rightarrow \text{tan } \theta \times \dfrac{1 - 2 + 2 \text{cos}^2 \theta}{2 \text{cos}^2 \theta - 1}

Notice that the numerator and the denominator are identical, which simplifies to:

\Rightarrow \text{tan } \theta \times \dfrac{2 \text{cos}^2 \theta - 1}{2 \text{cos}^2 \theta - 1}

This simplifies further to:

\Rightarrow \text{tan } \theta

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{sin } \theta - 2 \text{sin}^3 \theta}{2 \text{cos}^3 \theta - \text{ cos } \theta}$ = tan $\theta$ .

Question 24

Prove the following identities :

$\dfrac{\text{cos A}}{\text{1 - sin A}}$ = sec A + tan A

Answer:

To verify this identity, let’s start with the left-hand side:

\Rightarrow \dfrac{\text{cos A}}{\text{1 - sin A}}

Multiply both the numerator and the denominator by $(1 + \text{sin A})$ to simplify:

\Rightarrow \dfrac{\text{cos A(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}

Using the difference of squares identity, we know that $(1 - \text{sin A})(1 + \text{sin A})$ becomes $1 - \text{sin}^2 A$ . Thus, our expression becomes:

\Rightarrow \dfrac{\text{cos A(1 + sin A)}}{1 - \text{sin}^2 A}

Recall the Pythagorean identity $\text{cos}^2 A = 1 - \text{sin}^2 A$ to further simplify:

\Rightarrow \dfrac{\text{cos A(1 + sin A)}}{\text{cos}^2 A}

Breaking it down, we have:

\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}

This splits into two separate fractions:

\Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}

Recognizing these as trigonometric identities, we have:

\Rightarrow \text{sec A + tan A}

Thus, we find that the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{cos A}}{\text{1 - sin A}}$ = sec A + tan A.

Question 25

Prove the following identities :

$\dfrac{\text{sin A tan A}}{\text{1 - cos A}}$ = 1 + sec A

Answer:

Let’s work through the left-hand side of the equation:

\Rightarrow \dfrac{\text{sin A} \times \dfrac{\text{sin A}}{\text{cos A}}}{\text{1 - cos A}}

This simplifies to:

\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos A(1 - cos A)}}

Recall the Pythagorean identity:

\text{sin}^2 A = 1 - \text{cos}^2 A

Substituting this identity gives us:

\Rightarrow \dfrac{1 - \text{cos}^2 A}{\text{cos A(1 - cos A)}}

Now, factor the numerator:

\Rightarrow \dfrac{(1 - \text{cos A})(1 + \text{cos A})}{\text{cos A(1 - cos A)}}

Cancel out the common terms:

\Rightarrow \dfrac{1 + \text{cos A}}{\text{cos A}}

This can be split into two fractions:

\Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{cos A}}{\text{cos A}}

Simplifying gives:

\Rightarrow \text{sec A + 1}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{sin A tan A}}{\text{1 - cos A}}$ = 1 + sec A.

Question 26

Prove the following identities :

(1 + cot A – cosec A)(1 + tan A + sec A) = 2

Answer:

Let’s work through the left-hand side of the identity:

\Rightarrow \Big(1 + \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}\Big)\Big(1 + \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)

We simplify each term:

\Rightarrow \Big(\dfrac{\text{sin A + cos A - 1}}{\text{sin A}}\Big)\Big(\dfrac{\text{cos A + sin A + 1}}{\text{cos A}}\Big)

Now, multiply the numerators and the denominators:

\Rightarrow \dfrac{\text{(sin A + cos A - 1)(sin A + cos A + 1)}}{\text{sin A cos A}}

Expanding the numerator using the identity ((a-b)(a+b) = a^2 – b^2):

\Rightarrow \dfrac{\text{sin}^2 A + \text{sin A cos A + sin A + cos A sin A + cos A} + \text{ cos}^2 A - \text{sin A - cos A - 1}}{\text{sin A cos A}}

Combine like terms and apply the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ :

\Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + \text{2 cos A sin A - 1}}{\text{sin A cos A}}.

Substituting $\sin^2 A + \cos^2 A = 1$ gives:

\Rightarrow \dfrac{1 + \text{2 cos A sin A - 1}}{\text{sin A cos A}}

Simplify the expression:

\Rightarrow \dfrac{\text{2 cos A sin A}}{\text{cos A sin A}}

Cancel the common terms to get:

\Rightarrow 2.

Thus, the left-hand side equals the right-hand side.

Hence, proved that (1 + cot A – cosec A)(1 + tan A + sec A) = 2.

Question 27

Prove the following identities :

$\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A

Answer:

Let’s work through the left-hand side of the given identity:

\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}

To simplify, multiply both the numerator and the denominator by $\sqrt{1 + \text{sin A}}$ :

\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} \times \sqrt{\dfrac{\text{1 + sin A}}{\text{1 + sin A}}}

This results in:

\Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}}

Notice that the denominator becomes a difference of squares:

\Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{1 - sin}^2 \text{A}}}

Since $1 - \text{sin}^2 \text{A} = \text{cos}^2 \text{A}$ , we have:

\Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A}}

Taking the square root of the numerator and the denominator separately gives:

\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}}

This expression can be split into two fractions:

\Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}

Recognize these as the definitions of secant and tangent:

\Rightarrow \text{sec A + tan A}.

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A.

Question 28

Prove the following identities :

$\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A – cot A

Answer:

To demonstrate the equality, let’s start with the left-hand side:

\Rightarrow \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}

Multiply both the numerator and the denominator by $\sqrt{1 - \text{cos A}}$ :

\Rightarrow \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} \times \sqrt{\dfrac{\text{1 - cos A}}{\text{1 - cos A}}}

This simplifies to:

\Rightarrow \sqrt{\dfrac{\text{(1 - cos A)(1 - cos A)}}{\text{(1 + cos A)(1 - cos A)}}}

Which further simplifies to:

\Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{(1 - cos}^2 A)}}

Now, applying the identity $1 - \cos^2 A = \sin^2 A$ :

\Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{sin}^2 A}}

Simplifying the square root gives:

\Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A}}

This can be rewritten as:

\Rightarrow \dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}}

Which is equivalent to:

\Rightarrow \text{cosec A - cot A}.

Since the left-hand side equals the right-hand side, the identity is proven:

Hence, proved that $\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A – cot A.

Question 29

Prove the following identities :

1 – $\dfrac{\text{cos}^2 A}{\text{1 + sin A}}$ = sin A

Answer:

Consider the left-hand side of the equation:

\begin{aligned}\Rightarrow 1 - \dfrac{\text{cos}^2 A}{\text{1 + sin A}} \\\Rightarrow \dfrac{\text{1 + sin A - cos}^2 A}{\text{1 + sin A}}\end{aligned}

Recall the identity:

\text{cos}^2 A = 1 - \text{sin}^2 A

Substituting this in, we have:

\begin{aligned}\Rightarrow \dfrac{\text{1 + sin A - (1 - sin}^2 A)}{\text{1 + sin A}} \\\Rightarrow \dfrac{\text{1 + sin A - 1 + sin}^2 A}{\text{1 + sin A}} \\\Rightarrow \dfrac{\text{sin A + sin}^2 A}{\text{1 + sin A}}\end{aligned}

Notice that the numerator can be factored as:

\Rightarrow \dfrac{\text{sin A(1 + sin A)}}{\text{1 + sin A}}

Since the terms $\text{1 + sin A}$ cancel out, we are left with:

$\Rightarrow \text{sin A}$ .

∴ L.H.S. = R.H.S.

Hence, proved that 1 – $\dfrac{\text{cos}^2 A}{\text{1 + sin A}}$ = sin A.

Question 30

Prove the following identities :

\dfrac{1}{\text{sin A + cos A}} + \dfrac{1}{\text{sin A - cos A}} = \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}

Answer:

Let’s work on the left-hand side (L.H.S.) of the equation:

\Rightarrow \dfrac{\text{sin A - cos A + sin A + cos A}}{\text{(sin A + cos A)(sin A - cos A)}}

Notice that the numerator simplifies to:

\Rightarrow \dfrac{\text{2 sin A}}{\text{sin}^2 A - \text{cos}^2 A}

Now, recall the identity for $\text{sin}^2 A$ , which is $1 - \text{cos}^2 A$ . Applying this, we get:

\Rightarrow \dfrac{\text{2 sin A}}{\text{1 - \cos}^2 A - \text{cos}^2 A}

This further simplifies to:

\Rightarrow \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}

∴ The left-hand side equals the right-hand side (R.H.S.).

Hence, proved that $\dfrac{1}{\text{sin A + cos A}} + \dfrac{1}{\text{sin A - cos A}} = \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}$ .

Question 31

Prove the following identities :

\dfrac{\text{sin A + \text{cos A}}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}

Answer:

Let’s tackle the left-hand side of the identity:

\begin{aligned}\Rightarrow \dfrac{(\text{sin A + cos A})^2 + (\text{sin A - cos A})^2}{(\text{sin A - cos A})(\text{sin A + cos A})} \\\Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 2 \text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A - 2 \text{sin A cos A}}{\text{sin}^2 A - \text{cos}^2 A} \\\Rightarrow \dfrac{2(\text{sin}^2 A + \text{cos}^2 A)}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}

Recall the Pythagorean identity:

\text{sin}^2 A + \text{cos}^2 A = 1

Substituting this into our equation, we have:

\begin{aligned}\Rightarrow \dfrac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\\Rightarrow \dfrac{2}{\text{sin}^2 A - 1 + \text{sin}^2 A} \\\Rightarrow \dfrac{2}{2 \text{sin}^2 A - 1}\end{aligned}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{sin A + \text{cos A}}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{2 \text{sin}^2 A - 1}$ .

Question 32

Prove the following identities :

$\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A)

Answer:

Consider the left-hand side of the given identity:

$\Rightarrow \dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ .

To simplify, let’s combine the fractions:

$\Rightarrow \dfrac{\text{(1 + sin A)(cosec A + cot A)} - \text{(1 - sin A)(cosec A - cot A)}}{\text{cosec}^2 A - \text{cot}^2 A}$ .

Recall the identity $\text{cosec}^2 A - \text{cot}^2 A = 1$ .

Substituting this, we have:

$\Rightarrow (1 + \text{sin A})(\text{cosec A} + \text{cot A}) - (1 - \text{sin A})(\text{cosec A} - \text{cot A})$ .

Expand the terms:

$\Rightarrow \text{cosec A} + \text{cot A} + \text{sin A cosec A} + \text{sin A cot A} - (\text{cosec A} - \text{cot A} - \text{sin A cosec A} + \text{sin A cot A})$ .

Simplify by combining like terms:

$\Rightarrow \text{cosec A} - \text{cosec A} + \text{cot A} + \text{cot A} + \text{sin A cosec A} + \text{sin A cosec A} + \text{sin A cot A} - \text{sin A cot A}$ .

This reduces to:

$\Rightarrow 2 \text{cot A} + 2 \text{sin A cosec A}$ .

Recognize that $\text{sin A cosec A} = \text{sin A} \times \dfrac{1}{\text{sin A}} = 1$ :

$\Rightarrow 2 \text{cot A} + 2$ .

Factor out the common factor:

$\Rightarrow 2(\text{cot A} + 1)$ .

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + \text{cot A}).

Question 33

Prove the following identities :

$\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ – 1

Answer:

Let’s tackle the left-hand side of the identity:

\Rightarrow \dfrac{\text{cos θ} \times \dfrac{\text{cos θ}}{\text{sin θ}}}{\text{(1 + sin θ)}}

This simplifies to:

\Rightarrow \dfrac{\text{cos}^2 θ}{\text{sin θ(1 + sin θ)}}

Recall the identity for $\text{cos}^2 θ$ :

\text{cos}^2 θ = 1 - \text{sin}^2 θ

Substituting, we have:

\Rightarrow \dfrac{\text{1 - sin}^2 θ}{\text{sin θ(1 + sin θ)}}

Notice that the numerator is a difference of squares:

\Rightarrow \dfrac{\text{(1 - sin θ)(1 + sin θ)}}{\text{sin θ(1 + sin θ)}}

Cancel out the common factor $\text{(1 + sin θ)}$ :

\Rightarrow \dfrac{\text{1 - sin θ}}{\text{sin θ}}

This can be split into two fractions:

\Rightarrow \dfrac{1}{\text{sin θ}} - \dfrac{\text{sin θ}}{\text{sin θ}}

Which simplifies to:

\Rightarrow \text{cosec θ - 1}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ – 1.

Exercise 21(B)

Question 1(a)

(1 + cot^2 A) + (1 + tan^2 A) is equal to :

(a) $\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$
(b) sec^2 A – cosec^2 A
(c) sin^2 A – sin^4 A
(d) sec^2 A.cosec^2 A

Answer: (d) sec^2 A.cosec^2 A

We start with the expression:

⇒ $(1 + cot^2 A) + (1 + tan^2 A)$

Recalling the identity $1 + cot^2 A = cosec^2 A$ and $1 + tan^2 A = sec^2 A$ , we can rewrite:

⇒ $cosec^2 A + sec^2 A$

Expressing these in terms of sine and cosine:

⇒ $\dfrac{1}{\text{sin}^2 A} + \dfrac{1}{\text{cos}^2 A}$

Combine the fractions:

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A. \text{cos}^2 A}$

Using $\text{sin}^2 A + \text{cos}^2 A = 1$ , substitute to get:

⇒ $\dfrac{1}{\text{sin}^2 A. \text{cos}^2 A}$

This simplifies to:

⇒ $cosec^2 A. sec^2 A$

Thus, we conclude that the expression simplifies to $sec^2 A. cosec^2 A$ . Hence, Option 4 is the correct option.

Question 1(b)

If a = tan θ and b = sec θ, the relation between a and b is :

(a) a × b = 1
(b) a^2 – b^2 = 1
(c) b^2 – a^2 = 1
(d) a^2 + b^2 = 1

Answer: (c) b^2 – a^2 = 1

Consider substituting the given values: let $a = \tan \theta$ and $b = \sec \theta$ . Now, examine the expression for option 3:

b^2 - a^2

This translates to:

\sec^2 \theta - \tan^2 \theta

We know from trigonometric identities that $\sec^2 \theta - \tan^2 \theta = 1$ .

∴ The left-hand side equals the right-hand side.

Hence, Option 3 is the correct option.

Question 1(c)

$\dfrac{1}{\text{cot θ + tan θ}}$ is equal to :

(a) sin θ + cos θ
(b) sin θ. cos θ
(c) $\dfrac{1}{\text{sin θ. cos θ}}$
(d) $\dfrac{1}{\text{sin θ + cos θ}}$

Answer: (b) sin θ. cos θ

To find the value of $\dfrac{1}{\text{cot } \theta + \text{tan } \theta}$ , we start by expressing cotangent and tangent in terms of sine and cosine:

\Rightarrow \dfrac{1}{\dfrac{\text{cos } \theta}{\text{sin } \theta} + \dfrac{\text{sin } \theta}{\text{cos } \theta}}

Combine these fractions:

\Rightarrow \dfrac{1}{\dfrac{\text{cos}^2 \theta + \text{sin}^2 \theta}{\text{cos } \theta \cdot \text{sin } \theta}}

When we simplify, we get:

\Rightarrow \dfrac{\text{cos } \theta \cdot \text{sin } \theta}{\text{cos}^2 \theta + \text{sin}^2 \theta}

Recall that $\text{sin}^2 \theta + \text{cos}^2 \theta = 1$ .

Substitute this identity:

\Rightarrow \text{sin } \theta \cdot \text{cos } \theta

Hence, Option 2 is the correct option.

Question 1(d)

(sec θ – cos θ)^2 – (sec θ + cos θ)^2 is equal to :

(a) 4
(b) 2
(c) -2
(d) -4

Answer: (d) -4

Consider the expression ((\sec \theta – \cos \theta)^2 – (\sec \theta + \cos \theta)^2). We can apply the identity for the difference of squares: (a^2 – b^2 = (a + b)(a – b)).

⇒ ((\sec \theta – \cos \theta + \sec \theta + \cos \theta) $(\sec \theta - \cos \theta) - (\sec \theta + \cos \theta)$ )

Now, simplify each part:

⇒ ((\sec \theta – \cos \theta + \sec \theta + \cos \theta)) simplifies to $2 \sec \theta$ .

⇒ ((\sec \theta – \sec \theta – \cos \theta – \cos \theta)) simplifies to $-2 \cos \theta$ .

Combining these results:

⇒ (2 \sec \theta \times (-2 \cos \theta))

⇒ $-4 \sec \theta \cos \theta$

Since $\sec \theta = \frac{1}{\cos \theta}$ , we substitute:

⇒ $-4 \times \dfrac{1}{\cos \theta} \times \cos \theta$

⇒ $-4$ .

Thus, the final result is $-4$ . Hence, Option 4 is the correct option.

Question 1(e)

(cot A – cot B)^2 + (1 + cot A cot B)^2 is equal to :

(a) sec^2 A – cos^2 A
(b) sec^2 A – cosec^2 A
(c) (1 + tan A)^2 – (1 – cot A)^2
(d) cosec^2 A . cosec^2 B

Answer: (d) cosec^2 A . cosec^2 B

Consider the expression: $(\cot A - \cot B)^2 + (1 + \cot A \cot B)^2$ .

Let’s expand and simplify it:

⇒ $\cot^2 A + \cot^2 B - 2 \cot A \cot B + 1 + \cot^2 A \cdot \cot^2 B + 2 \cot A \cot B$

Notice that the terms $-2 \cot A \cot B$ and $+2 \cot A \cot B$ cancel each other out. So, we’re left with:

⇒ $\cot^2 A + \cot^2 A \cdot \cot^2 B + 1 + \cot^2 B$

Combine like terms:

⇒ $\cot^2 A(1 + \cot^2 B) + (1 + \cot^2 B)$

Factor out $(1 + \cot^2 B)$ :

⇒ $(1 + \cot^2 B)(1 + \cot^2 A)$

Now, express $\cot^2$ in terms of $\sin$ and $\cos$ :

⇒ $\Big(1 + \dfrac{\cos^2 B}{\sin^2 B}\Big)\Big(1 + \dfrac{\cos^2 A}{\sin^2 A}\Big)$

Using the identity $\sin^2 \theta + \cos^2 \theta = 1$ , rewrite it as:

⇒ $\Big(\dfrac{\sin^2 B + \cos^2 B}{\sin^2 B}\Big)\Big(\dfrac{\sin^2 A + \cos^2 A}{\sin^2 A}\Big)$

Since $\sin^2 \theta + \cos^2 \theta = 1$ , we substitute:

⇒ $\Big(\dfrac{1}{\sin^2 B}\Big)\Big(\dfrac{1}{\sin^2 A}\Big)$

This is the same as:

⇒ $\cosec^2 B \cdot \cosec^2 A$

Hence, option 4 is the correct option.

Question 2(i)

Prove that:

(sec A – tan A)^2 (1 + sin A) = (1 – sin A)

Answer:

We aim to demonstrate the equality:

(\sec A - \tan A)^2 (1 + \sin A) = (1 - \sin A)

Let’s work on the left-hand side (L.H.S.) of this equation:

\Rightarrow (\sec A - \tan A)^2 \text{(1 + sin A)}

Notice that $\sec A = \dfrac{1}{\cos A}$ and $\tan A = \dfrac{\sin A}{\cos A}$ . Substitute these identities in:

\Rightarrow \Big(\dfrac{1}{\cos A} - \dfrac{\sin A}{\cos A}\Big)^2 \text{(1 + sin A)}

Simplify the expression inside the brackets:

\Rightarrow \Big(\dfrac{1 - \sin A}{\cos A}\Big)^2 \text{(1 + sin A)}

Now, expand the square:

\Rightarrow \dfrac{(1 - \sin A)^2 (1 + \sin A)}{\cos^2 A}

Recall the Pythagorean identity $\cos^2 A = 1 - \sin^2 A$ . Replace $\cos^2 A$ :

\Rightarrow \dfrac{(1 - \sin A)^2 (1 + \sin A)}{1 - \sin^2 A}

Factor the denominator as a difference of squares:

\Rightarrow \dfrac{(1 - \sin A)^2 (1 + \sin A)}{(1 - \sin A)(1 + \sin A)}

Cancel the common terms $(1 + \sin A)$ in the numerator and denominator:

\Rightarrow \text{1 - sin A}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $(\sec A - \tan A)^2 (1 + \sin A) = (1 - \sin A)$ .

Question 2(ii)

Prove that :

$\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2

Answer:

To tackle the left-hand side of the equation, observe:

\Rightarrow \dfrac{(\text{cos}^3 A + \text{sin}^3 A)(\text{cos A - sin A}) + (\text{cos}^3 A - \text{sin}^3 A)(\text{cos A + sin A})}{\text{(cos A + sin A)(cos A - sin A)}}

Expanding the terms in the numerator gives:

\Rightarrow \dfrac{\text{cos}^4 A - \text{cos}^3 A\text{ sin A} + \text{cos A sin}^3 A - \text{ sin}^4 A + \text{cos}^4 A + \text{cos}^3 A\text{ sin A} - \text{ sin}^3 A\text{ cos A} - \text{sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A}

Notice that the middle terms cancel out, simplifying to:

\Rightarrow \dfrac{\text{2 cos}^4 A - \text{2 sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A}

Factor out the 2 and rewrite the expression:

\Rightarrow \dfrac{\text{2(cos}^4 A - \text{ sin}^4 A)}{\text{cos}^2 A - \text{sin}^2 A}

Recognize the difference of squares in the numerator:

\Rightarrow \dfrac{\text{2(cos}^2 A - \text{sin}^2 A)\text{(cos}^2 A + \text{sin}^2 A)}{\text{cos}^2 A - \text{sin}^2 A}

Cancel the common term $\text{cos}^2 A - \text{sin}^2 A$ in the numerator and denominator:

\Rightarrow \text{2 (cos}^2 A + \text{sin}^2 A).

Using the Pythagorean identity, $\text{cos}^2 A + \text{sin}^2 A = 1$ :

\Rightarrow 2 \times 1 = 2.

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2.

Question 2(iii)

Prove that :

$\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1

Answer:

To tackle the left-hand side of the expression, consider:

\begin{aligned}\Rightarrow \dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}} \\\Rightarrow \dfrac{\text{tan A}}{1 - \dfrac{1}{\text{tan A}}} + \dfrac{\dfrac{1}{\text{tan A}}}{\text{1 - tan A}} \\\Rightarrow \dfrac{\text{tan A}}{\dfrac{\text{tan A - 1}}{\text{tan A}}} + \dfrac{1}{\text{tan A(1 - tan A)}} \\\Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} + \dfrac{1}{\text{tan A(1 - tan A)}}\end{aligned}

Now, observe the subtraction:

\begin{aligned}\Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} - \dfrac{1}{\text{tan A(tan A - 1)}} \\\Rightarrow \dfrac{\text{tan}^3 A - 1}{\text{tan A(tan A - 1)}} \\\Rightarrow \dfrac{\text{(tan A - 1)(tan}^2 A + \text{ tan A + 1)}}{\text{tan A(tan A - 1)}} \\\Rightarrow \dfrac{\text{tan}^2 A + \text{ tan A + 1}}{\text{tan A}}\end{aligned}

Breaking it down further:

\begin{aligned}\Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A}} + \dfrac{\text{tan A}}{\text{tan A}} + \dfrac{1}{\text{tan A}} \\\Rightarrow \text{tan A + 1 + cot A} \\\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + \dfrac{\text{cos A}}{\text{sin A}} \\\Rightarrow \dfrac{\text{sin}^2 A + \text{sin A cos A + cos}^2 A}{\text{sin A cos A}}.\end{aligned}

Using the Pythagorean identity, we know:

\text{sin}^2 A + \text{cos}^2 A = 1

Therefore, substitute it back:

\begin{aligned}\Rightarrow \dfrac{1 + \text{sin A cos A}}{\text{sin A cos A}} \\\Rightarrow \dfrac{1}{\text{sin A cos A}} + \dfrac{\text{sin A cos A}}{\text{sin A cos A}} \\\Rightarrow \text{cosec A sec A} + 1.\end{aligned}

Thus, the left-hand side equals the right-hand side, confirming the identity:

Hence, proved that $\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1.

Question 2(iv)

Prove that :

\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)

Answer:

To tackle the left-hand side of the equation, observe:

\Rightarrow \Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2

Notice that $\text{tan A} = \dfrac{\text{sin A}}{\text{cos A}}$ . Substitute this into the expression:

\Rightarrow \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A}}{\text{cos A}} - \dfrac{1}{\text{cos A}}\Big)^2

Combine the fractions:

\Rightarrow \Big(\dfrac{\text{sin A + 1}}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A - 1}}{\text{cos A}}\Big)^2

Expand both squares:

\Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A}}{\text{cos}^2 A} + \dfrac{\text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A}

Combine the numerators over a common denominator:

\Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A} + \text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A}

Simplify the expression:

\Rightarrow \dfrac{\text{2(1 + sin}^2 A)}{\text{cos}^2 A}

Recall the identity $\text{cos}^2 A = 1 - \text{sin}^2 A$ . Substitute this identity:

\Rightarrow 2\Big(\dfrac{1 + \text{sin}^2 A}{1 - \text{sin}^2 A}\Big)

Since both sides of the equation are equal, we conclude:

Hence, proved that

$\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)$ .

Question 2(v)

Prove that :

2 sin^2 A + cos^4 A = 1 + sin^4 A

Answer:

Let’s work through the left-hand side of the equation:

⇒ 2 \sin^2 A + \cos^4 A

Notice that we can express \cos^4 A as (\cos^2 A)^2.

⇒ 2 \sin^2 A + (\cos^2 A)^2

Recall the identity for \cos^2 A, which is \cos^2 A = 1 – \sin^2 A.

⇒ 2 \sin^2 A + (1 – \sin^2 A)^2

Now, expand (1 – \sin^2 A)^2:

⇒ 2 \sin^2 A + 1 + \sin^4 A – 2 \sin^2 A

Notice that 2 \sin^2 A cancels out:

⇒ 1 + \sin^4 A.

Thus, the left-hand side equals the right-hand side, confirming that:

Hence, proved that 2 \sin^2 A + \cos^4 A = 1 + \sin^4 A.

Question 2(vi)

Prove that :

$\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0

Answer:

We start by examining the left-hand side of the equation:

\Rightarrow \dfrac{(\text{sin A - sin B})(\text{sin A + sin B}) + (\text{cos A - cos B})(\text{cos A + cos B})}{(\text{cos A + cos B})(\text{sin A + sin B})} = 0

First, expand the numerator:

\Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 B + \text{cos}^2 A - \text{cos}^2 B}{(\text{cos A + cos B})(\text{sin A + sin B})}

Rearrange terms using the identity $\text{sin}^2 \theta + \text{cos}^2 \theta = 1$ :

\Rightarrow \dfrac{(\text{sin}^2 A + \text{cos}^2 A) - (\text{sin}^2 B + \text{cos}^2 B)}{(\text{cos A + cos B})(\text{sin A + sin B})}

Apply the identity to simplify:

\therefore \dfrac{1 - 1}{(\text{cos A + cos B})(\text{sin A + sin B})}

This simplifies to:

\Rightarrow 0.

Thus, we have shown that the left-hand side equals the right-hand side. Hence, proved that $\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}} = 0$ .

Question 2(vii)

Prove that :

(cosec A – sin A)(sec A – cos A) = $\dfrac{1}{\text{tan A + cot A}}$

Answer:

To demonstrate the equality, let’s first handle the left-hand side:

\Rightarrow (\text{cosec A - sin A})(\text{sec A - cos A})

This can be rewritten using reciprocal identities:

\Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big)

Simplifying further, we have:

\Rightarrow \Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big)

Recall the Pythagorean identities:

$1 - \text{sin}^2 A = \text{cos}^2 A$
$1 - \text{cos}^2 A = \text{sin}^2 A$

Substituting these, we get:

\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}}

This simplifies to:

$\Rightarrow \text{cos A sin A}$ .

Next, let’s simplify the right-hand side:

\Rightarrow \dfrac{1}{\text{tan A + cot A}}

Expressing tan and cot in terms of sine and cosine gives us:

\Rightarrow \dfrac{1}{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}}

Combine the fractions in the denominator:

\Rightarrow \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}}

Using the identity $\text{sin}^2 \theta + \text{cos}^2 \theta = 1$ :

\Rightarrow \dfrac{\text{sin A cos A}}{1}

This simplifies to:

$\Rightarrow \text{sin A cos A}$ .

∴ L.H.S. = R.H.S.

Hence, proved that (cosec A – sin A)(sec A – cos A) = $\dfrac{1}{\text{tan A + cot A}}$ .

Question 2(viii)

Prove that :

(1 + tan A. tan B)^2 + (tan A – tan B)^2 = sec^2 A sec^2 B

Answer:

We start by recalling the identity:

\sec^2 \theta = 1 + \tan^2 \theta

Now, let’s simplify the left-hand side (LHS):

\Rightarrow (1 + \tan A \cdot \tan B)^2 + (\tan A - \tan B)^2

Expanding both squares, we have:

\Rightarrow 1 + \tan^2 A \tan^2 B + 2 \tan A \tan B + \tan^2 A + \tan^2 B - 2 \tan A \tan B

Notice how the $+ 2 \tan A \tan B$ and $- 2 \tan A \tan B$ cancel each other out:

\Rightarrow 1 + \tan^2 A \tan^2 B + \tan^2 A + \tan^2 B

Reorganizing terms gives:

\Rightarrow 1 + \tan^2 A + \tan^2 A \tan^2 B + \tan^2 B

Applying the identity $\sec^2 A = 1 + \tan^2 A$ , we substitute:

\Rightarrow \sec^2 A + \tan^2 B(\tan^2 A + 1)

Since $\tan^2 A + 1 = \sec^2 A$ , we substitute again:

\Rightarrow \sec^2 A + \tan^2 B \sec^2 A

Factor out $\sec^2 A$ :

\Rightarrow \sec^2 A(1 + \tan^2 B)

Using the identity once more, $1 + \tan^2 B = \sec^2 B$ , we get:

\Rightarrow \sec^2 A \sec^2 B.

Thus, the left-hand side equals the right-hand side.

Hence, proved that $(1 + \tan A \cdot \tan B)^2 + (\tan A - \tan B)^2 = \sec^2 A \sec^2 B$ .

Question 2(ix)

Prove that :

$\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A

Answer:

To tackle the left-hand side of the equation, consider:

\Rightarrow \dfrac{\text{cos A + sin A + 1 + cos A + sin A - 1}}{\text{(cos A + sin A - 1)(cos A + sin A + 1)}}

This simplifies to:

\Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{(cos A + sin A)}^2 - 1^2}

Using the identity $a^2 - b^2 = (a+b)(a-b)$ , rewrite the denominator:

\Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{cos}^2 A + \text{sin}^2 A + \text{2 cos A sin A} - 1}

Since $\text{cos}^2 A + \text{sin}^2 A = 1$ , this becomes:

\Rightarrow \dfrac{\text{2(cos A + sin A)}}{1 - 1 + \text{2 cos A sin A}}

Which further simplifies to:

\Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{2 cos A sin A}}

Cancel out the common factor of 2:

\Rightarrow \dfrac{\text{(cos A + sin A)}}{\text{cos A sin A}}

Separate the terms in the numerator:

\Rightarrow \dfrac{\text{cos A}}{\text{cos A sin A}} + \dfrac{\text{sin A}}{\text{cos A sin A}}

This gives:

\Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{1}{\text{cos A}}

Thus, we have:

\Rightarrow \text{cosec A + sec A}.

∴ L.H.S. = R.H.S., proving that $\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ equals cosec A + sec A.

Question 3

If x cos A + y sin A = m and x sin A – y cos A = n, then prove that :

x^2 + y^2 = m^2 + n^2

Answer:

We are given:

x \cos A + y \sin A = m \quad \text{and} \quad x \sin A - y \cos A = n

We need to show that:

x^2 + y^2 = m^2 + n^2

Let’s substitute the expressions for $m$ and $n$ in the right-hand side:

= (x \cos A + y \sin A)^2 + (x \sin A - y \cos A)^2

Expanding both squares, we have:

= x^2 \cos^2 A + y^2 \sin^2 A + 2xy \cos A \sin A + x^2 \sin^2 A + y^2 \cos^2 A - 2xy \sin A \cos A

Notice that the middle terms cancel each other out:

= x^2 \cos^2 A + x^2 \sin^2 A + y^2 \cos^2 A + y^2 \sin^2 A

We can factor the terms:

= x^2(\sin^2 A + \cos^2 A) + y^2(\sin^2 A + \cos^2 A)

Using the identity $\sin^2 A + \cos^2 A = 1$ , we simplify further:

= x^2 \times 1 + y^2 \times 1

= x^2 + y^2

Thus, the left-hand side equals the right-hand side, confirming that:

Hence, proved that $x^2 + y^2 = m^2 + n^2$ .

Question 4

If m = a sec A + b tan A and n = a tan A + b sec A, then prove that :

m^2 – n^2 = a^2 – b^2

Answer:

We need to demonstrate that $m^2 - n^2 = a^2 - b^2$ .

First, substitute the expressions for $m$ and $n$ into the left-hand side:

(a \sec A + b \tan A)^2 - (a \tan A + b \sec A)^2

Expanding these squares, we have:

a^2 \sec^2 A + b^2 \tan^2 A + 2ab \sec A \tan A - (a^2 \tan^2 A + b^2 \sec^2 A + 2ab \sec A \tan A)

Notice that the terms $2ab \sec A \tan A$ cancel each other out, leaving:

a^2 \sec^2 A - a^2 \tan^2 A + b^2 \tan^2 A - b^2 \sec^2 A

This can be rearranged as:

a^2 (\sec^2 A - \tan^2 A) + b^2 (\tan^2 A - \sec^2 A)

Recognizing that $\sec^2 A - \tan^2 A = 1$ from trigonometric identities, we simplify:

a^2 \times 1 - b^2 \times 1

Thus, we arrive at:

a^2 - b^2

Therefore, the left-hand side equals the right-hand side. Hence, proved that $m^2 - n^2 = a^2 - b^2$ .

Question 5

If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that :

x^2 + y^2 + z^2 = r^2

Answer:

We aim to establish that:

⇒ x^2 + y^2 + z^2 = r^2

Start by substituting the expressions for x, y, and z into the left side of the equation:

= (r \sin A \cos B)^2 + (r \sin A \sin B)^2 + (r \cos A)^2

This expands to:

= r^2 \sin^2 A \cos^2 B + r^2 \sin^2 A \sin^2 B + r^2 \cos^2 A

Notice that we can factor out r^2 \sin^2 A from the first two terms:

= r^2 \sin^2 A (\cos^2 B + \sin^2 B) + r^2 \cos^2 A

Using the identity \sin^2 \theta + \cos^2 \theta = 1, simplify further:

⇒ r^2 \sin^2 A + r^2 \cos^2 A

Factor r^2 from the entire expression:

⇒ r^2 (\sin^2 A + \cos^2 A)

Again, using \sin^2 \theta + \cos^2 \theta = 1, we find:

⇒ r^2 \times 1

⇒ r^2.

Therefore, the left-hand side equals the right-hand side.

Hence, proved that x^2 + y^2 + z^2 = r^2.

Question 6

If $\dfrac{\text{cos A}}{\text{cos B}} = m \text{ and } \dfrac{\text{cos A}}{\text{sin B}}$ = n,

show that:

(m^2 + n^2) cos^2 B = n^2

Answer:

We need to establish that

(m^2 + n^2) \text{cos}^2 B = n^2

Let’s substitute the given expressions for $m$ and $n$ into the left-hand side of the equation:

$\Rightarrow \Big$ \Big(\dfrac{\text{cos A}}{\text{cos B}}\Big)^2 + \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2\Big $\text{cos}^2 B$

This simplifies to:

$\Rightarrow \Big$ \dfrac{\text{cos}^2 A}{\text{cos}^2 B} + \dfrac{\text{cos}^2 A}{\text{sin}^2 B}\Big $\text{cos}^2 B$

Now, combine the fractions:

\Rightarrow \dfrac{\text{cos}^2 A \text{sin}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{cos}^2 B \text{sin}^2 B} \times \text{cos}^2 B

Factor out $\text{cos}^2 A$ :

\Rightarrow \dfrac{\text{cos}^2 A(\text{sin}^2 B + \text{cos}^2 B)}{\text{sin}^2 B}

Recall the Pythagorean identity $\text{sin}^2 B + \text{cos}^2 B = 1$ :

\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 B}

Recognizing this as:

\Rightarrow \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2

This is simply $n^2$ .

Since the left-hand side equals the right-hand side, we have:

Hence, proved that $(m^2 + n^2) \text{cos}^2 B = n^2$ .

Exercise 21(C)

Question 1(a)

sin^2 A + sin^2 (90° – A) is equal to :

(a) -1
(b) 1
(c) 2
(d) -2

Answer: (b) 1

We know that $\sin (90^\circ - A) = \cos A$ .

We need to evaluate the expression:

\sin^2 A + \sin^2 (90^\circ - A)

Substituting the identity, we have:

\sin^2 A + \cos^2 A

According to the Pythagorean identity, $\sin^2 A + \cos^2 A = 1$ .

Hence, Option 2 is the correct option.

Question 1(b)

In a triangle ABC, sec $\dfrac{C + A}{2}$ is equal to :

(a) cosec $\dfrac{B}{2}$
(b) sec $\dfrac{B}{2}$
(c) cosec $\dfrac{B + A}{2}$
(d) none of these

Answer: (a) cosec

\dfrac{B}{2}

Consider triangle ABC. Using the angle sum property, we know:

∴ A + B + C = 180°

From this, deduce that:

∴ A + C = 180° – B

Now, dividing both sides by 2, we have:

∴ $\dfrac{A + C}{2} = \dfrac{180° - B}{2}$ ……..(1)

Substitute this result into sec $\dfrac{C + A}{2}$ :

\Rightarrow \text{sec } \dfrac{C + A}{2}

\Rightarrow \text{sec } \dfrac{180° - B}{2}

\Rightarrow \text{sec } \Big(90° - \dfrac{B}{2}\Big)

The secant of $90° - \theta$ is cosecant $\theta$ , thus:

\Rightarrow \text{cosec } \dfrac{B}{2}

Hence, option 1 is the correct option.

Question 1(c)

$\dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}}$ + 5 is equal to :

(a) -3
(b) 4
(c) 3
(d) -4

Answer: (c) 3

To find the value of the given expression, start by simplifying each term:

\Rightarrow \dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}} + 5

Notice that ( \text{cot 46°} = \text{cot (90° – 44°)} ) and ( \text{sec 20°} = \text{sec (90° – 70°)} ). Using these identities, the expression becomes:

\Rightarrow \dfrac{\text{tan 44°}}{\text{tan 44°}} - 3 \dfrac{\text{cosec 70°}}{\text{cosec 70°}} + 5

Since $\dfrac{\text{tan 44°}}{\text{tan 44°}} = 1$ and $\dfrac{\text{cosec 70°}}{\text{cosec 70°}} = 1$ , the expression simplifies to:

\Rightarrow 1 - 3 \times 1 + 5

This further simplifies to:

\Rightarrow 1 - 3 + 5

Calculating the above gives:

\Rightarrow 3.

Hence, option 3 is the correct option.

Question 1(d)

sin 67° . cos 23° + cos 67° . sin 23° is equal to :

(a) -1
(b) 2 sin 67°
(c) 2 cos 23°
(d) 1

Answer: (d) 1

Consider the expression:

⇒ $\sin 67^\circ \cdot \cos 23^\circ + \cos 67^\circ \cdot \sin 23^\circ$

Notice that $23^\circ$ and $67^\circ$ add up to $90^\circ$ . Therefore, we can rewrite $\cos 23^\circ$ as $\cos (90^\circ - 67^\circ)$ and $\sin 23^\circ$ as $\sin (90^\circ - 67^\circ)$ .

⇒ $\sin 67^\circ \cdot \cos (90^\circ - 67^\circ) + \cos 67^\circ \cdot \sin (90^\circ - 67^\circ)$

Using the identity $\cos(90^\circ - \theta) = \sin \theta$ and $\sin(90^\circ - \theta) = \cos \theta$ , the expression becomes:

⇒ $\sin 67^\circ \cdot \sin 67^\circ + \cos 67^\circ \cdot \cos 67^\circ$

⇒ $\sin^2 67^\circ + \cos^2 67^\circ$

According to the Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$ .

⇒ $1$

Hence, Option 4 is the correct option.

Question 1(e)

$\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}$ is equivalent to :

(a) cos^2 θ – 1
(b) sin^2 θ
(c) sin^2 θ – cos^2 θ
(d) sin^2 θ – 1

Answer: (b) sin^2 θ

To simplify the expression, start with:

\dfrac{\text{cos } \theta \cdot \text{cos } (90^\circ - \theta)}{\text{tan } (90^\circ - \theta)}

Notice that ( \text{cos } (90^\circ – \theta) = \text{sin } \theta ) and ( \text{tan } (90^\circ – \theta) = \text{cot } \theta ). Substituting these identities, the expression becomes:

\dfrac{\text{cos } \theta \cdot \text{sin } \theta}{\text{cot } \theta}

Since $\text{cot } \theta = \dfrac{\text{cos } \theta}{\text{sin } \theta}$ , replace $\text{cot } \theta$ in the denominator:

\dfrac{\text{cos } \theta \cdot \text{sin } \theta}{\dfrac{\text{cos } \theta}{\text{sin } \theta}}

Multiply by the reciprocal:

\dfrac{\text{cos } \theta \cdot \text{sin } \theta \cdot \text{sin } \theta}{\text{cos } \theta}

The $\text{cos } \theta$ in the numerator and denominator cancel out, leaving:

\text{sin}^2 \theta

Thus, option 2 is the correct option.

Question 2(i)

Show that :

tan 10° tan 15° tan 75° tan 80° = 1

Answer:

Consider the left-hand side expression:

⇒ tan 10° tan 15° tan 75° tan 80°

Recognize that 75° and 80° can be rewritten using complementary angles:

⇒ tan 10° tan 15° tan (90° – 15°) tan (90° – 10°)

We know from trigonometric identities that:

\tan(90° - A) = \cot A

Applying this identity, we have:

⇒ tan 10° tan 15° cot 15° cot 10°

This simplifies to:

⇒ tan 10° × tan 15° × $\dfrac{1}{\text{tan 15°}}$ × $\dfrac{1}{\text{tan 10°}}$

Notice how the terms cancel each other out:

⇒ 1.

∴ the left-hand side equals the right-hand side.

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

Question 2(ii)

Show that :

sin 42° sec 48° + cos 42° cosec 48° = 2

Answer:

Let’s evaluate the left-hand side of the expression:

⇒ ( \sin 42^\circ \sec (90^\circ – 42^\circ) + \cos 42^\circ \cosec (90^\circ – 42^\circ) )

Utilizing the complementary angle identities:

( \sec (90^\circ – A) = \cosec A )
( \cosec (90^\circ – A) = \sec A )

This transforms the expression into:

⇒ $\sin 42^\circ \cosec 42^\circ + \cos 42^\circ \sec 42^\circ$

Breaking it down further:

⇒ $\sin 42^\circ \times \dfrac{1}{\sin 42^\circ} + \cos 42^\circ \times \dfrac{1}{\cos 42^\circ}$

Simplifying gives:

⇒ 1 + 1

⇒ 2.

Therefore, the left-hand side equals the right-hand side.

Hence, proved that $\sin 42^\circ \sec 48^\circ + \cos 42^\circ \cosec 48^\circ = 2$ .

Question 3

Express each of the following in terms of angles between 0° and 45° :

(i) sin 59° + tan 63°

(ii) cosec 68° + cot 72°

Answer:

(i) Let’s tackle the first expression:

⇒ sin 59° + tan 63°

Notice that 59° can be written as (90° – 31°) and 63° as (90° – 27°). Thus,

⇒ sin (90° – 31°) + tan (90° – 27°)

Using the identities, $\sin (90° - A) = \cos A$ and $\tan (90° - A) = \cot A$ ,

⇒ \cos 31° + \cot 27°.

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

(ii) Now, for the second expression:

⇒ cosec 68° + cot 72°

We can express 68° as (90° – 22°) and 72° as (90° – 18°), leading to:

⇒ cosec (90° – 22°) + cot (90° – 18°)

Applying the identities, $\cosec (90° - A) = \sec A$ and $\cot (90° - A) = \tan A$ ,

⇒ \sec 22° + \tan 18°.

Hence, cosec 68° + cot 72° = sec 22° + tan 18°.

Question 4(i)

Show that :

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

Answer:

We aim to demonstrate that:

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ equals sec A cosec A.

Consider the trigonometric identities:

$\text{sin(90° - A)} = \text{cos A}$
$\text{cos(90° - A)} = \text{sin A}$

Substitute these identities into the left-hand side:

\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}

Combine the terms over a common denominator:

\Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}

Recall the Pythagorean identity:

\text{sin}^2 A + \text{cos}^2 A = 1

Substituting this into our expression gives:

\Rightarrow \dfrac{1}{\text{sin A cos A}}

This can be rewritten as:

\Rightarrow \dfrac{1}{\text{sin A}} \times \dfrac{1}{\text{cos A}}

Which simplifies to:

$\Rightarrow \text{cosec A sec A}$ .

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A.

Question 4(ii)

Show that :

sin A cos A – $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0

Answer:

Recall the trigonometric identities:

$\cos (90^\circ - A) = \sin A$
$\sec (90^\circ - A) = \cosec A$
$\cosec (90^\circ - A) = \sec A$
$\sin (90^\circ - A) = \cos A$

Using these, we can rewrite the expression as:

\Rightarrow \sin A \cos A - \dfrac{\sin A \sin A \cos A}{\cosec A} - \dfrac{\cos A \cos A \sin A}{\sec A}

Simplify further:

\Rightarrow \sin A \cos A - \dfrac{\sin^2 A \cos A}{\dfrac{1}{\sin A}} - \dfrac{\cos^2 A \sin A}{\dfrac{1}{\cos A}}

This simplifies to:

\Rightarrow \sin A \cos A - \sin^3 A \cos A - \cos^3 A \sin A

Factor out $\sin A \cos A$ :

\Rightarrow \sin A \cos A - \sin A \cos A (\sin^2 A + \cos^2 A)

Using the identity $\sin^2 A + \cos^2 A = 1$ :

\Rightarrow \sin A \cos A - \sin A \cos A

This results in:

\Rightarrow 0.

Hence, proved that $\sin A \cos A - \dfrac{\sin A \cos (90^\circ - A) \cos A}{\sec (90^\circ - A)} - \dfrac{\cos A \sin (90^\circ - A) \sin A}{\cosec (90^\circ - A)} = 0$ .

Question 5

For triangle ABC, show that :

(i) sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$

(ii) tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$

Answer:

(i) Consider triangle ABC:

∴ ∠A + ∠B + ∠C = 180° (using the angle sum property of a triangle)

∴ ∠A + ∠B = 180° – ∠C ……….(1)

We need to show:

sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$

Substitute the expression for (A + B) from equation (1) into the left side:

$\Rightarrow \text{sin} \dfrac{180° - C}{2}$ \
$\Rightarrow \text{sin} \Big(90° - \dfrac{C}{2}\Big)$

Recall the identity:

sin(90° – θ) = cos θ

∴ $\text{cos} \dfrac{C}{2}$ .

Thus, the left side equals the right side.

Therefore, sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$ .

(ii) Again, in triangle ABC:

∴ ∠A + ∠B + ∠C = 180° (by the angle sum property)

∴ ∠B + ∠C = 180° – ∠A ……….(1)

We need to demonstrate:

tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$

Replace (B + C) from equation (1) in the left side:

$\Rightarrow \text{tan} \dfrac{180° - A}{2}$ \
$\Rightarrow \text{tan } \Big(90° - \dfrac{A}{2}\Big)$

Recall the identity:

tan(90° – θ) = cot θ

$\Rightarrow \text{cot} \dfrac{A}{2}$ .

Thus, the left side matches the right side.

Therefore, tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$ .

Question 6(i)

Evaluate :

3 cos 80° cosec 10° + 2 sin 59° sec 31°

Answer:

Let’s evaluate the expression:

⇒ 3 \text{ cos } 80° \text{ cosec } 10° + 2 \text{ sin } 59° \text{ sec } 31°

Observe the angles: 10° and 80° are complementary (adding up to 90°), as are 59° and 31°. Thus, we can use complementary angle identities:

(\text{cosec}(90° – \theta) = \text{sec } \theta)
(\text{sec}(90° – \theta) = \text{cosec } \theta)

Applying these identities:

⇒ 3 \text{ cos } 80° \text{ sec } 80° + 2 \text{ sin } 59° \text{ cosec } 59°

Now simplify using the reciprocal identities:

⇒ 3 \text{ cos } 80° \times \frac{1}{\text{cos } 80°} + 2 \text{ sin } 59° \times \frac{1}{\text{sin } 59°}

The terms simplify to:

⇒ 3 + 2

⇒ 5.

Thus, the value of the expression is 5.

Question 6(ii)

Evaluate :

$\dfrac{\text{sin 80°}}{\text{cos 10°}}$ + sin 59° sec 31°

Answer:

To solve this expression, we start with:

\Rightarrow \dfrac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°}

Notice that $80°$ can be rewritten as $90° - 10°$ , and $31°$ can be expressed as $90° - 59°$ . This allows us to use the complementary angle identities:

\Rightarrow \dfrac{\text{sin (90 - 10)°}}{\text{cos 10°}} + \text{sin 59° sec (90 - 59)°}

Recall the identities: $\text{sin(90° - θ)} = \text{cos θ}$ and $\text{sec(90° - θ)} = \text{cosec θ}$ .

Applying these, we get:

\Rightarrow \dfrac{\text{cos 10°}}{\text{cos 10°}} + \text{sin 59° cosec 59°}

The first term simplifies to $1$ because $\dfrac{\text{cos 10°}}{\text{cos 10°}} = 1$ . For the second term, $\text{sin 59°}$ multiplied by $\dfrac{1}{\text{sin 59°}}$ also simplifies to $1$ .

Thus, the expression becomes:

\Rightarrow 1 + 1

Which equals $2$ .

Hence, $\dfrac{\text{sin 80°}}{\text{cos 10°}}$ + sin 59° sec 31° = 2.

Question 6(iii)

Evaluate :

tan(55° – A) – cot(35° + A)

Answer:

To find the value of the expression, we start with:

⇒ ( \tan(55° – A) – \cot(35° + A) )

Notice that ( \tan(55° – A) ) can be rewritten using the complementary angle identity:

⇒ ( \tan[90° – (35° + A)] – \cot(35° + A) )

According to the identity ( \tan(90° – \theta) = \cot \theta ), we have:

⇒ ( \cot(35° + A) – \cot(35° + A) )

The expression simplifies to zero:

⇒ $0$ .

Hence, ( \tan(55° – A) – \cot(35° + A) = 0 )

Question 6(iv)

Evaluate :

cosec (65° + A) – sec (25° – A)

Answer:

To solve the expression, we start with:

⇒ cosec (65° + A) – sec (25° – A)

Notice that we can rewrite the first term using the identity for cosecant:

⇒ cosec [90° – (25° – A)] – sec (25° – A)

According to the trigonometric identity:

cosec(90° – θ) = sec θ

This simplifies our expression to:

⇒ sec (25° – A) – sec (25° – A)

Since both terms are identical, their difference is:

⇒ 0.

Hence, cosec (65° + A) – sec (25° – A) = 0.

Question 6(v)

Evaluate :

$2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - \sqrt{2}$ cos 45°

Answer:

Let’s simplify the expression step by step:

\Rightarrow 2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - \sqrt{2}\text{cos 45°}

Notice that $\text{cot 33°} = \text{tan (90° - 33°)}$ and $\text{tan 20°} = \text{cot (90° - 20°)}$ . Applying these identities, we have:

\Rightarrow 2\dfrac{\text{tan 57°}}{\text{tan (90° - 57°)}} - \dfrac{\text{cot 70°}}{\text{cot (90° - 70°)}} - \sqrt{2}\times \dfrac{1}{\sqrt{2}}

Given the trigonometric identities $\text{tan(90° - θ)} = \text{cot θ}$ and $\text{cot(90° - θ)} = \text{tan θ}$ , the expression becomes:

\Rightarrow 2 \times \dfrac{\text{tan 57°}}{\text{tan 57°}} - \dfrac{\text{cot 70°}}{\text{cot 70°}} - 1

Simplifying further, we get:

\Rightarrow 2 - 1 - 1

\Rightarrow 0.

Thus, the evaluated expression is 0.

Question 6(vi)

Evaluate :

\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}

Answer:

We start with the expression:

\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}

Notice that $49°$ is the complement of $41°$ , i.e., $49° = 90° - 41°$ . Similarly, $15°$ is the complement of $75°$ . Using the complementary angle identities, we have:

$\tan (90° - \theta) = \cot \theta$
$\cos (90° - \theta) = \sin \theta$

Applying these identities, the expression becomes:

\dfrac{\text{cot}^2 41°}{\text{tan}^2 (90° - 41°)} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 (90° - 75°)}

Substituting the identities:

\Rightarrow \dfrac{\text{cot}^2 41°}{\text{cot}^2 41°} - 2\dfrac{\text{sin}^2 75°}{\text{sin}^2 75°}

This simplifies to:

\Rightarrow 1 - 2

Thus, the final result is:

\Rightarrow -1.

Hence, $\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}$ = -1

Question 6(vii)

\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30°

Answer:

Let’s explore the expression:

\Rightarrow \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30°

Notice how we can rewrite the sines using the identity $\sin (90° - \theta) = \cos \theta$ :

\Rightarrow \dfrac{\text{cos 70°}}{\text{sin (90° - 70°)}} + \dfrac{\text{cos 59°}}{\text{sin (90° - 59°)}} - \text{8 sin}^2 30°

Applying this identity, we have:

\Rightarrow \dfrac{\text{cos 70°}}{\text{cos 70°}} + \dfrac{\text{cos 59°}}{\text{cos 59°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2

Here, the fractions simplify to 1 each, and $\sin 30°$ is $\dfrac{1}{2}$ , so:

\Rightarrow 1 + 1 - 8 \times \dfrac{1}{4}

Calculating further, we get:

\Rightarrow 2 - 2

Thus, the expression simplifies to:

\Rightarrow 0.

Hence, $\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30°$ = 0.

Question 7

A triangle ABC is right angled at B; find the value of $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$

Answer:

In the right-angled triangle ABC, where ∠B = 90°, we know:

∠A + ∠B + ∠C = 180°.

Substituting ∠B = 90°, we have:

∠A + 90° + ∠C = 180°

This simplifies to:

∠A + ∠C = 90°

So, ∠A = 90° – ∠C.

Now, substitute ∠A in the expression $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ :

\Rightarrow \dfrac{\text{sec (90° - C). cosec C - tan (90° - C). cot C}}{\text{sin 90°}}

Using trigonometric identities, we know:

$\tan (90° - C) = \cot C$
$\sec (90° - C) = \cosec C$
$\cosec^2 C - \cot^2 C = 1$

Substitute these identities into the expression:

\Rightarrow \dfrac{\text{cosec C. cosec C - cot C. cot C}}{1}

This simplifies to:

\Rightarrow \text{cosec}^2 C - \text{cot}^2 C

Which equals:

\Rightarrow 1.

Therefore, $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ = 1.

Question 8(i)

Find value of x, if :

sin x = sin 60° cos 30° + cos 60° sin 30°

Answer:

To find the value of $x$ , let’s evaluate the given expression:

\Rightarrow \sin x = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2}

Notice that:

$\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{3}{4}$
$\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

Combining these, we have:

\Rightarrow \sin x = \dfrac{3}{4} + \dfrac{1}{4}

This simplifies to:

\Rightarrow \sin x = \dfrac{4}{4}

Which further reduces to:

\Rightarrow \sin x = 1

Since $\sin 90^\circ = 1$ , it follows that:

\Rightarrow x = 90^\circ

Hence, $x = 90^\circ$ .

Question 8(ii)

Find value of x, if :

cos x = cos 60° cos 30° – sin 60° sin 30°

Answer:

To find the value of $x$ , we start with the given identity:

\Rightarrow \cos x = \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ

Substitute the known trigonometric values:

\Rightarrow \cos x = \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}

Calculate the expression on the right side:

\Rightarrow \cos x = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}

Notice that the terms cancel each other out:

\Rightarrow \cos x = 0

We know that $\cos 90^\circ = 0$ , thus:

\Rightarrow \cos x = \cos 90^\circ

This implies:

\Rightarrow x = 90^\circ.

Hence, x = 90°.

Question 8(iii)

Find value of x, if :

tan x = $\dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan 60° tan 30°}}$

Answer:

Let’s work through the calculation step by step. We start with the expression:

\Rightarrow \text{tan x} = \dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan 60° tan 30°}}

First, substitute the known values for tan 60° and tan 30°:

\Rightarrow \text{tan x} = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}}

Simplify the expression in the numerator and denominator:

\Rightarrow \text{tan x} = \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1}

This further simplifies to:

\Rightarrow \text{tan x} = \dfrac{\dfrac{2}{\sqrt{3}}}{2}

Now, simplify the fraction:

\Rightarrow \text{tan x} = \dfrac{1}{\sqrt{3}}

Recognizing that $\dfrac{1}{\sqrt{3}}$ is the same as $\text{tan 30°}$ , we find:

\Rightarrow \text{tan x} = \text{tan 30°}

Thus, it follows that:

\Rightarrow x = 30°

Hence, x = 30°.

Question 8(iv)

Find value of x, if :

sin 3x = 2 sin 30° cos 30°

Answer:

To find the value of $x$ , we start with the equation:

⇒ $\sin 3x = 2 \sin 30° \cos 30°$

Substitute the known values for $\sin 30°$ and $\cos 30°$ :

⇒ $\sin 3x = 2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}$

Simplifying the right side gives:

⇒ $\sin 3x = \dfrac{\sqrt{3}}{2}$

We know that $\sin 60° = \dfrac{\sqrt{3}}{2}$ , so:

⇒ $\sin 3x = \sin 60°$

This implies:

⇒ $3x = 60°$

Dividing both sides by 3, we find:

⇒ $x = \dfrac{60°}{3} = 20°$

Hence, $x = 20°$ .

Question 8(v)

Find value of x, if :

cos(2x – 6°) = cos^2 30° – cos^2 60°

Answer:

To find the value of $x$ , we start with the equation:

⇒ $\cos(2x - 6°) = \cos^2 30° - \cos^2 60°$

Calculate each cosine squared value:

⇒ $\cos(2x - 6°) = \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2$

Simplify the squares:

⇒ $\cos(2x - 6°) = \dfrac{3}{4} - \dfrac{1}{4}$

Subtract the fractions:

⇒ $\cos(2x - 6°) = \dfrac{2}{4}$

Simplify the fraction:

⇒ $\cos(2x - 6°) = \dfrac{1}{2}$

Recognize that $\dfrac{1}{2}$ is $\cos 60°$ :

⇒ $\cos(2x - 6°) = \cos 60°$

Since the cosine values are equal, equate the angles:

⇒ $2x - 6° = 60°$

Solve for $2x$ :

⇒ $2x = 66°$

Finally, divide by 2 to find $x$ :

⇒ $x = 33°$ .

Hence, $x = 33°$ .

Question 9

In each case given below, find the value of angle A, where 0° ≤ A ≤ 90°.

(i) sin (90° – 3A). cosec 42° = 1

(ii) cos (90° – A). sec 77° = 1

Answer:

(i) We have the equation:

⇒ ( \sin (90° – 3A) \cdot \text{cosec } 42° = 1 )

Using the identity for cosecant, $\text{cosec } 42° = \dfrac{1}{\sin 42°}$ , we can rewrite the equation as:

⇒ ( \sin (90° – 3A) \times \dfrac{1}{\sin 42°} = 1 )

This simplifies to:

⇒ ( \sin (90° – 3A) = \sin 42° )

Since the sine values are equal, we equate the angles:

⇒ $90° - 3A = 42°$

Solving for $A$ , we get:

⇒ $3A = 90° - 42°$

⇒ $3A = 48°$

⇒ $A = 16°$ .

Hence, $A = 16°$ .

(ii) For the second equation:

⇒ ( \cos (90° – A) \cdot \text{sec } 77° = 1 )

Using the identity for secant, $\text{sec } 77° = \dfrac{1}{\cos 77°}$ , rewrite the equation as:

⇒ ( \cos (90° – A) \times \dfrac{1}{\cos 77°} = 1 )

This simplifies to:

⇒ ( \cos (90° – A) = \cos 77° )

Equating the angles, we have:

⇒ $90° - A = 77°$

Solving for $A$ , we find:

⇒ $A = 90° - 77°$

⇒ $A = 13°$ .

Hence, $A = 13°$ .

Question 10(i)

Prove that :

$\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 – cos^2 θ

Answer:

We know from trigonometric identities that:

cos(90° – θ) = sin θ.

Now, let’s work through the left-hand side of the equation:

\Rightarrow \dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}

Using the identity for cos(90° – θ), substitute sin θ:

\Rightarrow \dfrac{\text{sin θ cos θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}}

Simplify the expression by multiplying the numerator by the reciprocal of the denominator:

\Rightarrow \text{sin}^2 θ

Recall that sin²θ + cos²θ = 1, which implies sin²θ = 1 – cos²θ:

\Rightarrow \text{1 - cos}^2 θ.

∴ L.H.S. = R.H.S., confirming our proof.

Hence, proved that $\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 – cos^2 θ.

Question 11

Evaluate :

\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}

Answer:

Let’s simplify the expression:

\begin{aligned}\Rightarrow \dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°} \\\Rightarrow \dfrac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 10° - \text{tan}^2 (90° - 10°)}\end{aligned}

Notice that we can use the co-function identities:

$\text{sin}(90° - \theta) = \text{cos} \theta$
$\text{cos}(90° - \theta) = \text{sin} \theta$
$\text{tan}(90° - \theta) = \text{cot} \theta$

Applying these, we have:

\begin{aligned}\Rightarrow \dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{cosec}^2 10° - \text{cot}^2 10°} \\\Rightarrow \dfrac{\text{sin}^2 35° + \text{cos}^2 35°}{\text{cosec}^2 10° - \text{cot}^2 10°}\end{aligned}

Now, use the Pythagorean identities:

$\text{sin}^2 \theta + \text{cos}^2 \theta = 1$
$\text{cosec}^2 \theta - \text{cot}^2 \theta = 1$

Thus, the expression simplifies to:

\Rightarrow \dfrac{1}{1}

⇒ 1.

Hence, $\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}$ = 1.

Question 12

Evaluate :

sin^2 34° + sin^2 56° + 2 tan 18° tan 72° – cot^2 30°

Answer:

Let’s evaluate the expression step by step:

\Rightarrow \sin^2 34^\circ + \sin^2 56^\circ + 2 \tan 18^\circ \tan 72^\circ - \cot^2 30^\circ

Notice that $56^\circ$ is the complement of $34^\circ$ , and $72^\circ$ is the complement of $18^\circ$ . Thus, we can rewrite using complementary angle identities:

\Rightarrow \sin^2 34^\circ + \sin^2 (90^\circ - 34^\circ) + 2 \tan 18^\circ \tan (90^\circ - 18^\circ) - \cot^2 30^\circ

Using the identities $\sin(90^\circ - A) = \cos A$ and $\tan(90^\circ - A) = \cot A$ , we have:

\Rightarrow \sin^2 34^\circ + \cos^2 34^\circ + 2 \tan 18^\circ \cot 18^\circ - \cot^2 30^\circ

Applying the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ :

\Rightarrow 1 + 2 \tan 18^\circ \times \dfrac{1}{\tan 18^\circ} - (\sqrt{3})^2

Simplifying further, we get:

\Rightarrow 1 + 2 - 3

Therefore:

\Rightarrow 3 - 3

Thus, the final result is:

\Rightarrow 0.

Hence, $\sin^2 34^\circ + \sin^2 56^\circ + 2 \tan 18^\circ \tan 72^\circ - \cot^2 30^\circ = 0$ .

Question 13

Evaluate :

cosec^2 57° – tan^2 33° + cos 44° cosec 46° – $\sqrt{2}$ cos 45° – tan^2 60°

Answer:

Let’s tackle the given expression step by step:

⇒ $\text{cosec}^2 57° - \text{tan}^2 33° + \text{cos} 44° \cdot \text{cosec} 46° - \sqrt{2} \cdot \text{cos} 45° - \text{tan}^2 60°$

Notice that $\text{tan}^2 33°$ can be rewritten using the complementary angle identity: ( \text{tan}(90° – 57°) = \text{cot} 57° ). Similarly, ( \text{cosec}(90° – 44°) = \text{sec} 44° ).

⇒ ( \text{cosec}^2 57° – \text{cot}^2 57° + \text{cos} 44° \cdot \text{sec} 44° – \sqrt{2} \cdot \frac{1}{\sqrt{2}} – (\sqrt{3})^2 )

Applying the identity $\text{cosec}^2 A - \text{cot}^2 A = 1$ , we have:

⇒ $1 + \text{cos} 44° \times \frac{1}{\text{cos} 44°} - 1 - 3$

Simplifying further:

⇒ $1 + 1 - 1 - 3$

⇒ $-2$ .

Hence, $\text{cosec}^2 57° - \text{tan}^2 33° + \text{cos} 44° \cdot \text{cosec} 46° - \sqrt{2} \cdot \text{cos} 45° - \text{tan}^2 60° = -2$ .

Exercise 21(D)

Question 1(a)

If 2 sin θ = $\sqrt{3}$ , then :

(a) 2 cos θ + 1 = 0
(b) 1 – sin θ = 0
(c) cos θ = $\sqrt{3}$
(d) 2 cos θ = 1

Answer: (d) 2 cos θ = 1

We start with the given equation:

2 \sin \theta = \sqrt{3}

Dividing both sides by 2, we find:

\sin \theta = \frac{\sqrt{3}}{2}

Recognizing this value, we know that:

\sin \theta = \sin 60^\circ

Thus, $\theta = 60^\circ$ .

Next, we determine the cosine of $\theta$ :

\cos \theta = \cos 60^\circ

Now, substituting $\cos 60^\circ = \frac{1}{2}$ into the expression:

2 \cos \theta = 2 \times \frac{1}{2}

This simplifies to:

2 \cos \theta = 1

Hence, Option 4 is the correct option.

Question 1(b)

If cos 62° 40′ = x, the value of 1 + sin^2 27° 20′ is :

(a) 1 + x^2
(b) 1 – x^2
(c) x^2 – 1
(d) 1 + x

Answer: (a) 1 + x^2

We have that $\cos 62^\circ 40' = x$ . Notice that $62^\circ 40'$ is the complement of $27^\circ 20'$ , since $90^\circ - 27^\circ 20' = 62^\circ 40'$ . Thus, ( \cos(90^\circ – 27^\circ 20′) = x ) implies $\sin 27^\circ 20' = x$ .

Now, substituting $\sin 27^\circ 20' = x$ into the expression $1 + \sin^2 27^\circ 20'$ , we find:

1 + \sin^2 27^\circ 20' = 1 + x^2.

Hence, Option 1 is the correct option.

Question 1(c)

sin θ = 0.5 implies tan θ + cot θ is equal to :

(a) 2 : $\sqrt{3}$
(b) $\sqrt{3}$ : 4
(c) 4 : $\sqrt{3}$
(d) $\sqrt{3}$ : 2

Answer: (c) 4 :

\sqrt{3}

We know that:

∴ sin θ = 0.5

This can be rewritten as:

∴ sin θ = \dfrac{5}{10}

⇒ sin θ = \dfrac{1}{2}

Recognizing that \dfrac{1}{2} is the sine of 30°, we have:

∴ θ = 30°

Now, let’s find the value of tan θ + cot θ:

⇒ tan θ + cot θ

Substituting θ = 30°, we get:

⇒ tan 30° + cot 30°

Using the trigonometric values, we find:

⇒ \dfrac{1}{\sqrt{3}} + \sqrt{3}

Combining these into a single fraction:

⇒ \dfrac{1 + 3}{\sqrt{3}}

Simplifying gives us:

⇒ \dfrac{4}{\sqrt{3}}

Expressing this as a ratio:

⇒ 4 : \sqrt{3}

Hence, the expression tan θ + cot θ evaluates to 4 : \sqrt{3}.

Therefore, Option 3 is the correct option.

Question 1(d)

If cos 63° 36′ = 0.4446, the value of sin 26° 24′ is :

(a) 1
(b) 1 + 0.4446
(c) 0.4446
(d) 1 – 0.4446

Answer: (c) 0.4446

We know that cos 63° 36′ = 0.4446.

Notice that 63° 36′ and 26° 24′ are complementary angles because their sum is 90°. Therefore, we can express this as:

⇒ cos (90° – 26° 24′) = 0.4446

Using the identity that cos(90° – θ) = sin θ, we can rewrite this as:

⇒ sin 26° 24′ = 0.4446

Hence, Option 3 is the correct option.

Question 2

Use tables to find sine of :

(i) 21°

(ii) 34° 42′

(iii) 47° 32′

(iv) 62° 57′

Answer:

(i) Referring to the trigonometric tables, we find:

sin 21° = 0.3584

Hence, sin 21° = 0.3584

(ii) According to the table values:

sin 34° 42′ = 0.5693

Hence, sin 34° 42′ = 0.5693

(iii) Using the tables, we have:

sin 47° 30′ = 0.7373

The difference for 2′ is 0.0004, which we need to add.

∴ sin 47° 32′ = 0.7373 + 0.0004 = 0.7377

Hence, sin 47° 32′ = 0.7377

(iv) From the tables, we get:

sin 62° 54′ = 0.8902

The difference for 3′ is 0.0004, which needs to be added.

∴ sin 62° 57′ = 0.8902 + 0.0004 = 0.8906

Hence, sin 62° 57′ = 0.8906

Question 3

Use tables to find cosine of :

(i) 2° 4′

(ii) 8° 12′

(iii) 26° 32′

(iv) 65° 41′

Answer:

(i) According to the table, the cosine of 2° 6′ is 0.9993. Since the difference for 2′ is 0, we simply add this to get:

∴ cos 2° 4′ = 0.9993 + 0 = 0.9993

Hence, 2° 4′ = 0.9993

(ii) From the table, we find:

cos 8° 12′ = 0.9898

Hence, 8° 12′ = 0.9898

(iii) The table shows that cos 26° 30′ is 0.8949. The difference for 2′ is 0.0003, which we subtract:

∴ cos 26° 32′ = 0.8949 – 0.0003 = 0.8946

Hence, 26° 32′ = 0.8946

(iv) Looking at the table, cos 65° 42′ is 0.4115. For a difference of 1′, we add 0.0003:

∴ cos 65° 41′ = 0.4115 + 0.0003 = 0.4118

Hence, 65° 41′ = 0.4118

Question 4

Use trigonometric tables to find tangent of :

(i) 37°

(ii) 42° 18′

Answer:

(i) Referring to the trigonometric table, the value for $\tan 37^\circ$ is found to be 0.7536.

Thus, $\tan 37^\circ = 0.7536$

(ii) When we look up the trigonometric table for $\tan 42^\circ 18'$ , the value is 0.9099.

Thus, $\tan 42^\circ 18' = 0.9099$

Question 5

Use tables to find the acute angle θ, if the value of sin θ is :

(i) 0.4848

(ii) 0.3827

Answer:

(i) Assume that $\sin \theta = 0.4848$ .

Consulting the trigonometric tables, we find:

\sin 29^\circ = 0.4848

∴ $\theta = 29^\circ$ .

Hence, $\theta = 29^\circ$ .

(ii) Assume that $\sin \theta = 0.3827$ .

According to the trigonometric graph,

\sin 22^\circ 30' = 0.3827

∴ $\theta = 22^\circ 30'$ .

Hence, $\theta = 22^\circ 30'$ .

Question 6

Use tables to find the acute angle θ, if the value of cos θ is :

(i) 0.9848

(ii) 0.9574

Answer:

(i) Suppose $\cos \theta = 0.9848$ .

Upon consulting the trigonometric tables, we find that $\cos 10^\circ = 0.9848$ .

Thus, $\theta = 10^\circ$ .

Hence, $\theta = 10^\circ$ .

(ii) Suppose $\cos \theta = 0.9574$ .

Checking the trigonometric tables reveals that $\cos 16^\circ 48' = 0.9573$ .

The difference between $\cos \theta$ and $\cos 16^\circ 48'$ is:

\cos \theta - \cos 16^\circ 48' = 0.9574 - 0.9573 = 0.0004

According to the tables, a difference of 1′ corresponds to 0.0004. Therefore,

\theta = 16^\circ 48' - 1' = 16^\circ 47'

Hence, $\theta = 16^\circ 47'$ .

Question 7

Use tables to find the acute angle θ, if the value of tan θ is :

(i) 0.2419

(ii) 0.4741

Answer:

(i) Consider that $\tan \theta = 0.2419$ .

Using the trigonometric tables, we find that:

\tan 13^\circ 36' = 0.2419

Therefore, $\theta = 13^\circ 36'$ .

Hence, $\theta = 13^\circ 36'$ .

(ii) Suppose $\tan \theta = 0.4741$ .

Referring to the tables, we observe:

\tan 25^\circ 18' = 0.4727

Calculating the difference, $\tan \theta - \tan 25^\circ 18' = 0.4741 - 0.4727 = 0.0014$ .

Consulting the tables again, we see that a difference of $4'$ corresponds to $0.0014$ .

∴ $\theta = 25^\circ 18' + 4' = 25^\circ 22'$ .

Hence, $\theta = 25^\circ 22'$ .

Test Yourself

Question 1(a)

For the given figure, which of the following conditions is true :

cot γ < cot β
cot γ > cot α
cot β > cot α
cot β < cot γ

Answer:

Observing the figure, we have the following expressions for the cotangent of the angles:

For $\alpha$ , $\cot \alpha = \dfrac{OA}{AB}$
For $\beta$ , $\cot \beta = \dfrac{OA}{AC}$
For $\gamma$ , $\cot \gamma = \dfrac{OA}{AD}$

Notice that the side lengths satisfy the inequality $AB < AC < AD$ . This implies that:

\frac{OA}{AD} < \frac{OA}{AC} < \frac{OA}{AB}

Thus, we conclude that $\cot \gamma < \cot \beta < \cot \alpha$ .

Hence, Option 1 is the correct option.

Question 1(b)

If sin A = $\dfrac{4}{5}$ , the value of $\sqrt{\dfrac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}}$ is :

(a) 1
(b) 3
(c) $\dfrac{1}{2}$
(d) $\dfrac{1}{3}$

Answer: (d)

\dfrac{1}{3}

We have $\sin A = \dfrac{4}{5}$ . Our task is to determine the value of the expression:

\sqrt{\dfrac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}}

Recall that $\cos(90° - A) = \sin A$ . Thus, we can rewrite the expression as:

\Rightarrow \sqrt{\dfrac{1 - \sin A}{1 + \sin A}}

Substitute $\sin A = \dfrac{4}{5}$ :

\Rightarrow \sqrt{\dfrac{1 - \dfrac{4}{5}}{1 + \dfrac{4}{5}}}

Simplify the fractions:

\Rightarrow \sqrt{\dfrac{\dfrac{5 - 4}{5}}{\dfrac{5 + 4}{5}}}

\Rightarrow \sqrt{\dfrac{\dfrac{1}{5}}{\dfrac{9}{5}}}

Multiply the numerator and denominator by 5 to eliminate the fractions:

\Rightarrow \sqrt{\dfrac{1 \times 5}{5 \times 9}}

\Rightarrow \sqrt{\dfrac{1}{9}}

The square root of $\dfrac{1}{9}$ is:

$\Rightarrow \dfrac{1}{3}$ .

Hence, option 4 is the correct option.

Question 1(c)

(1 + tan^2 A) × cos^2 A – 1 is :

(a) 1
(b) -2
(c) 0
(d) 2

Answer: (c) 0

To find the value of the expression, start by using the identity $1 + \text{tan}^2 A = \text{sec}^2 A$ . Thus, we have:

$\Rightarrow (1 + \text{tan}^2 A) \times \text{cos}^2 A - 1 \[1em] \Rightarrow \Big(1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1$ .

Now, notice that:

$\Rightarrow \Big(\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1$ .

Since $\text{cos}^2 A + \text{sin}^2 A = 1$ , the expression simplifies to:

$\Rightarrow \dfrac{1}{\text{cos}^2 A} \times \text{cos}^2 A - 1 \[1em] \Rightarrow 1 - 1 \[1em] \Rightarrow 0$ .

Hence, Option 3 is the correct option.

Question 1(d)

$\dfrac{\text{sin A tan A}}{1 - \text{cos}^2 A}$ is equal to :

(a) cos A
(b) sec A
(c) cosec A
(d) sin A

Answer: (b) sec A

Let’s simplify the expression:

\Rightarrow \dfrac{\text{sin A tan A}}{1 - \text{cos}^2 A}

Notice that $1 - \text{cos}^2 A$ can be rewritten using the Pythagorean identity $\text{sin}^2 A$ . Thus, the expression becomes:

\Rightarrow \dfrac{\text{sin A tan A}}{\text{sin}^2 A}

This simplifies to:

\Rightarrow \dfrac{\text{tan A}}{\text{sin A}}

Now, replace $\text{tan A}$ with $\dfrac{\text{sin A}}{\text{cos A}}$ :

\Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}}}{\text{sin A}}

Simplifying further, we have:

\Rightarrow \dfrac{\text{sin A}}{\text{sin A} \times \text{cos A}}

Cancel $\text{sin A}$ from the numerator and denominator:

\Rightarrow \dfrac{1}{\text{cos A}}

This is equal to $\text{sec A}$ .

Hence, Option 2 is the correct option.

Question 1(e)

For acute angles α and β, if α > β, then :

A : cos α > cos β

B : tan α > tan β

C : sin α > sin β

D : cot α > cot β

Which of the above statement/statements is/are true :

(a) A and B
(b) B and C
(c) C and D
(d) D and A

Answer: (b) B and C

When dealing with acute angles, it’s important to remember that as the angle increases, both the sine and tangent values increase as well.

Given that:

α > β

This implies that:

∴ $\tan \alpha > \tan \beta$

∴ $\sin \alpha > \sin \beta$

Hence, Option 2 is the correct option.

Question 1(f)

If M = x cos A + y sin A and N = x sin A – y cos A, then M^2 + N^2 is :

(a) x^2 – y^2
(b) y^2 – x^2
(c) x^2 + y^2
(d) (x + y)^2

Answer: (c) x^2 + y^2

Consider the expressions for $M$ and $N$ : $M = x \cos A + y \sin A$ and $N = x \sin A - y \cos A$ . We need to find $M^2 + N^2$ .

Starting with $M^2 + N^2$ :

(x \cos A + y \sin A)^2 + (x \sin A - y \cos A)^2

Expanding both squares, we get:

x^2 \cos^2 A + y^2 \sin^2 A + 2xy \sin A \cos A + x^2 \sin^2 A + y^2 \cos^2 A - 2xy \sin A \cos A

Notice that $+ 2xy \sin A \cos A$ and $- 2xy \sin A \cos A$ cancel each other out. Thus, we have:

x^2 \cos^2 A + y^2 \sin^2 A + x^2 \sin^2 A + y^2 \cos^2 A

Rearranging terms:

x^2 (\cos^2 A + \sin^2 A) + y^2 (\sin^2 A + \cos^2 A)

We know from the Pythagorean identity that $\sin^2 A + \cos^2 A = 1$ . Therefore, substituting this identity:

x^2 \times 1 + y^2 \times 1

This simplifies to:

x^2 + y^2

Thus, the value of $M^2 + N^2$ is $x^2 + y^2$ . Option 3 is the correct option.

Question 1(g)

$\dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}}$ is equal to :

(a) tan 30°
(b) tan 45°
(c) tan 60°
(d) none

Answer: (a) tan 30°

Consider the expression:

\dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}}

We know that $\text{tan 60°} = \sqrt{3}$ and $\text{tan 30°} = \dfrac{1}{\sqrt{3}}$ . Substituting these values, we have:

\Rightarrow \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}}

Simplifying the numerator and denominator:

\Rightarrow \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1}

This becomes:

\Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{2}

Simplifying further, we get:

\Rightarrow \dfrac{2}{2\sqrt{3}}

Which reduces to:

\Rightarrow \dfrac{1}{\sqrt{3}}

We recognize that $\dfrac{1}{\sqrt{3}}$ is equal to $\text{tan 30°}$ .

Hence, Option 1 is the correct option.

Question 1(h)

$3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45°$ is equal to :

(a) 0
(b) -1
(c) 1
(d) 2

Answer: (c) 1

To find the value of the expression, we proceed as follows:

3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45°

Notice that ( \text{cos 43°} = \text{cos (90° – 47°)} = \text{sin 47°} ) and ( \text{cot 40°} = \text{cot (90° – 50°)} = \text{tan 50°} ). Substituting these identities, we have:

3\dfrac{\text{sin 47°}}{\text{sin 47°}} - \dfrac{\text{tan 50°}}{\text{tan 50°}} - 2 \times \left(\dfrac{1}{\sqrt{2}}\right)^2

This simplifies to:

3 \times 1 - 1 - 2 \times \dfrac{1}{2}

Calculating further, we get:

3 - 1 - 1

Finally, the result is:

1

Hence, Option 3 is the correct option.

Question 1(i)

If sin 2x = 2 sin 45° cos 45°; the value of x is :

(a) 45°
(b) 90°
(c) 30°
(d) 60°

Answer: (a) 45°

To determine the value of $x$ , we start with the equation:

⇒ $\sin 2x = 2 \sin 45^\circ \cos 45^\circ$

Using the known values for $\sin 45^\circ$ and $\cos 45^\circ$ , we have:

⇒ $\sin 2x = 2 \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}}$

Simplifying the right side, we find:

⇒ $\sin 2x = 2 \times \dfrac{1}{2}$

⇒ $\sin 2x = 1$

Since $\sin 90^\circ = 1$ , we equate:

⇒ $\sin 2x = \sin 90^\circ$

This implies:

⇒ $2x = 90^\circ$

Dividing both sides by 2, we get:

⇒ $x = \dfrac{90^\circ}{2} = 45^\circ$

Hence, Option 1 is the correct option.

Question 1(j)

For acute angle θ, sec^2θ = 1 – tan^2θ

Assertion(A): sec^2θ = 1 – tan^2θ is not a trigonometric identity.

Reason(R): For an acute angle θ, the trigonometric equation is an identity, if it is satisfied for every value of angle θ.

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

The accurate trigonometric identity is given by:

sec^2θ = 1 + tan^2θ

This confirms that the assertion (A) is indeed true.

A trigonometric identity is defined as an equation that remains valid for every possible value of its variable, which in this context is the angle θ.

Thus, the reason (R) is also true.

∴ Both A and R are true, and R correctly explains A.

Hence, option 3 is the correct option.

Question 1(k)

(1 + tan^2 A)(1 – sin A)(1 + sin A)

Assertion(A): The value of given trigonometric expression is 0.

Reason(R): The given expression is equal to sec^2 A.cos^2 A

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (b) A is false, R is true.

Consider the expression:

⇒ ((1 + \tan^2 A)(1 – \sin A)(1 + \sin A))

Recognize that ((1 – \sin^2 A)) is equivalent to $\cos^2 A$ . Thus, the expression becomes:

⇒ ((1 + \tan^2 A)\cos^2 A)

Using the identity $1 + \tan^2 A = \sec^2 A$ , substitute to get:

⇒ $\sec^2 A \cdot \cos^2 A$

This simplifies to:

⇒ $\dfrac{1}{\cos^2 A} \cdot \cos^2 A$

⇒ 1.

The reason (R) is indeed true as the expression simplifies to $\sec^2 A \cdot \cos^2 A$ . However, the assertion (A) stating that the value is 0 is incorrect.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Question 1(l)

cos A = $\dfrac{\sqrt{3}}{2}$ and sin B = $\dfrac{1}{2}$

Assertion (A): tan (A + B) = $\sqrt{3}$

Reason (R): cos A = $\dfrac{\sqrt{3}}{2}$ ⇒ A = 30°

sin B = $\dfrac{1}{2}$ ⇒ B = 30°

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

We have that $\cos A = \dfrac{\sqrt{3}}{2}$ and $\sin B = \dfrac{1}{2}$ .

This implies $\cos A = \cos 30^\circ$ and $\sin B = \sin 30^\circ$ .

Thus, $A = 30^\circ$ and $B = 30^\circ$ .

Consequently, the reason (R) is valid.

Now, let’s calculate $\tan(A + B)$ :

$\tan(A + B) = \tan(30^\circ + 30^\circ) = \tan 60^\circ = \sqrt{3}$ .

Therefore, $\tan (A + B) = \sqrt{3}$ , confirming that the assertion (A) is true.

∴ Both the assertion (A) and reason (R) are true, and the reason (R) correctly explains the assertion (A).

Hence, option 3 is the correct option.

Question 1(m)

x = (cosec A + cot A)(1 – cos A)

Assertion (A): x = sin A

Reason (R): x = $\Big(\dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A}) = \dfrac{\text{sin}^2 A}{\text{sin A}} = \text{sin A}$

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

We start with the expression given for $x$ :

$x = (\text{cosec } A + \text{cot } A)(1 - \text{cos } A)$ .

Now, let’s express cosec and cot in terms of sin and cos:

$\Rightarrow x = \left(\dfrac{1}{\text{sin } A} + \dfrac{\text{cos } A}{\text{sin } A}\right)(1 - \text{cos } A)$ .

Combine the terms inside the bracket:

$= \left(\dfrac{1 + \text{cos } A}{\text{sin } A}\right)(1 - \text{cos } A)$ .

Apply the identity for the difference of squares:

$= \dfrac{(1 + \text{cos } A)(1 - \text{cos } A)}{\text{sin } A}$ .

This simplifies to:

$= \dfrac{1 - \text{cos}^2 A}{\text{sin } A}$ .

Using the Pythagorean identity $1 - \text{cos}^2 A = \text{sin}^2 A$ , we have:

$= \dfrac{\text{sin}^2 A}{\text{sin } A}$ .

Simplifying the fraction gives us:

$= \text{sin } A$ .

∴ Both the assertion and reason are true, and the reason correctly explains the assertion.

Hence, option 3 is the correct option.

Question 1(n)

$\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A – tan A

Statement (1): $\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A – tan A

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Statement (2): $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (c) Statement 1 is true, and statement 2 is false.

Given that $\dfrac{1 - \text{sin A}}{\text{cos A}} = \text{sec A} - \text{tan A}$ .

Let’s examine the left-hand side:

$\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}} = \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}} = \text{sec A} - \text{tan A}$ .

Since the left-hand side equals the right-hand side, the equation holds true.

Now consider $\dfrac{1 + \text{sin A}}{\text{cos A}} = \text{sec A} + \text{tan A}$ .

For the left-hand side:

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}} = \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} = \text{sec A} + \text{tan A}$ .

Here, the left-hand side matches the right-hand side, confirming the equation is correct.

Thus, Statement 1 is true.

Next, evaluate $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}} = 2\text{sec A}$ .

For the left-hand side:

\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}

\Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}} - \left(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\right)

\Rightarrow (\text{sec A} - \text{tan A}) - (\text{sec A} + \text{tan A})

\Rightarrow \text{sec A} - \text{tan A} - \text{sec A} - \text{tan A}

$\Rightarrow -2\text{tan A}$ .

Since the left-hand side does not equal the right-hand side, Statement 2 is false.

∴ Statement 1 is true and Statement 2 is false.

Hence, option 3 is the correct option.

Question 1(o)

$\text{cos}^2 θ + \dfrac{1}{1 + {\text{cot}^2 θ}}$ = x

Statement (1): x = 1

Statement (2): x = $\text{cos}^2 θ + \dfrac{1}{{\text{cosec}^2 θ}}$ = cos^2 θ + sin^2 θ

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a) Both the statement are true.

We start with the expression $\text{cos}^2 θ + \dfrac{1}{1 + {\text{cot}^2 θ}}$ and set it equal to $x$ .

Now, we know that $1 + \text{cot}^2 θ = \text{cosec}^2 θ$ . Substituting this identity into our expression, we have:

x = \text{cos}^2 θ + \dfrac{1}{\text{cosec}^2 θ}

Next, recall that $\dfrac{1}{\text{cosec}^2 θ} = \text{sin}^2 θ$ . Thus, the expression simplifies to:

x = \text{cos}^2 θ + \text{sin}^2 θ

According to the Pythagorean identity, $\text{cos}^2 θ + \text{sin}^2 θ = 1$ . Therefore:

x = 1

Thus, both statements are indeed true. Hence, option 1 is the correct option.

Question 1(p)

$\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ}$ = x

Statement (1): x = 1

Statement (2): x = $\dfrac{\text{sin}^2 θ - 1}{{\text{cos}^2 θ}} = \dfrac{-\text{cos}^2 θ}{{\text{cos}^2 θ}}$ = -1

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

We start with the expression $\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{\text{cos}^2 θ}$ and set it equal to $x$ .

⇒ $x = \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{\text{cos}^2 θ}$

This can be rewritten as:

⇒ $x = \dfrac{\text{sin}^2 θ - 1}{\text{cos}^2 θ}$

Recall the identity $\text{sin}^2 θ + \text{cos}^2 θ = 1$ . Therefore, $1 = \text{sin}^2 θ + \text{cos}^2 θ$ . Substituting this into our expression gives:

⇒ $x = \dfrac{\text{sin}^2 θ - (\text{sin}^2 θ + \text{cos}^2 θ)}{\text{cos}^2 θ}$

Simplifying the numerator, we have:

⇒ $x = \dfrac{\text{sin}^2 θ - \text{sin}^2 θ - \text{cos}^2 θ}{\text{cos}^2 θ}$

⇒ $x = \dfrac{- \text{cos}^2 θ}{\text{cos}^2 θ}$

⇒ $x = -1$

Thus, Statement 1 is incorrect, and Statement 2 is accurate.

Hence, option 4 is the correct option.

Question 1(q)

sin A = cos A and tan^2 A + cot^2 A + 2

Statement (1): A = 45°

tan^2 A + cot^2 A + 2 = 4

Statement (2): sin A = cos A ⇒ A = 45°

(a) Both the statement are true.
(b) Both the statement are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a) Both the statement are true.

The condition given is $\sin A = \cos A$ .

⇒ $\dfrac{\text{sin A}}{\text{cos A}} = 1$

⇒ $\tan A = 1$

⇒ $\tan A = \tan 45^\circ$

⇒ $A = 45^\circ$

This confirms that statement 2 is accurate.

Now, substituting $A = 45^\circ$ into $\tan^2 A + \cot^2 A + 2$ , we calculate:

⇒ $\tan^2 A + \cot^2 A + 2 = \tan^2 45^\circ + \cot^2 45^\circ + 2$

= $(1)^2 + (1)^2 + 2$

= $1 + 1 + 2$

= $4$ .

This verifies that statement 1 is also correct.

∴ Both statements are true.

Hence, option 1 is the correct option.

Question 2(i)

Prove the following identities :

\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}

Answer:

We start by focusing on the left-hand side of the identity:

\Rightarrow \dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}}

Combining the fractions, we have:

\Rightarrow \dfrac{\text{cos A - sin A + cos A + sin A}}{\text{cos}^2 A - \text{sin}^2 A}

Simplifying the numerator:

\Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - \text{sin}^2 A}

Recall the identity for $\text{sin}^2 A$ :

\text{sin}^2 A = 1 - \text{cos}^2 A

Substitute this into the denominator:

\Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - (1 - \text{cos}^2 A)}

Simplifying further:

\Rightarrow \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}

Thus, the left-hand side equals the right-hand side, confirming the identity.

Hence, proved that $\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}$ .

Question 2(ii)

Prove the following identities :

1 - \dfrac{\text{sin}^2 A}{\text{1 + cos A}} = \text{cos A}

Answer:

We start by recalling the identity:

\text{sin}^2 A = 1 - \text{cos}^2 A

Now, let’s focus on the left-hand side (L.H.S.) of the given identity:

\Rightarrow 1 - \dfrac{1 - \text{cos}^2 A}{1 + \text{cos A}}

Notice that we can factor the numerator:

\Rightarrow 1 - \dfrac{(1 - \text{cos A})(1 + \text{cos A})}{1 + \text{cos A}}

Here, the $(1 + \text{cos A})$ terms in the numerator and the denominator cancel out:

\Rightarrow 1 - (1 - \text{cos A})

Simplifying further gives:

\Rightarrow 1 - 1 + \text{cos A}

This reduces to:

$\Rightarrow \text{cos A}$ .

Thus, we have shown that the left-hand side equals the right-hand side.

Hence, proved that $1 - \dfrac{\text{sin}^2 A}{1 + \text{cos A}} = \text{cos A}$ .

Question 2(iii)

Prove the following identities :

\dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A = sec A}

Answer:

Let’s tackle the left-hand side of the identity:

\Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A}

We can express $\text{tan A}$ as $\dfrac{\text{sin A}}{\text{cos A}}$ . Thus, the expression becomes:

\Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{sin A}}{\text{cos A}}

Combine these fractions by finding a common denominator:

\Rightarrow \dfrac{\text{cos}^2 A + \text{sin A(1 + sin A)}}{\text{cos A(1 + sin A)}}

Simplify the numerator:

\Rightarrow \dfrac{\text{cos}^2 A + \text{sin A + sin}^2 A}{\text{cos A(1 + sin A)}}

Recall the Pythagorean identity $\text{cos}^2 A + \text{sin}^2 A = 1$ :

\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A(1 + sin A)}}

Cancel $\text{1 + sin A}$ from the numerator and denominator:

\Rightarrow \dfrac{1}{\text{cos A}}

This simplifies to $\text{sec A}$ .

Since the left-hand side equals the right-hand side, the identity is proven:

Hence, proved that $\dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A = sec A}$ .

Question 2(iv)

Prove the following identities :

\dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A = cosec A}

Answer:

Let’s tackle the left-hand side of the identity:

\Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A}

This can be rewritten as:

\Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} - \dfrac{\text{cos A}}{\text{sin A}}

Combine the fractions over a common denominator:

\Rightarrow \dfrac{\text{sin}^2 A - \text{cos A(1 - cos A)}}{\text{sin A(1 - cos A)}}

Simplifying the numerator gives:

\Rightarrow \dfrac{\text{sin}^2 A - \text{cos A + cos}^2 A}{\text{sin A(1 - cos A)}}

Recall the Pythagorean identity:

\text{sin}^2 A + \text{cos}^2 A = 1

Substitute into the fraction:

\Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A(1 - cos A)}}

Cancel the common term $(1 - \text{cos A})$ in the numerator and denominator:

\Rightarrow \dfrac{1}{\text{sin A}}

This simplifies to:

\Rightarrow \text{cosec A}

Thus, we have shown that the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A = cosec A}$ .

Question 2(v)

Prove the following identities :

\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}}

Answer:

To simplify the left-hand side of the equation, multiply both the numerator and denominator by $\sqrt{1 + \text{cos A}}$ :

\Rightarrow \sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} \times \sqrt{\dfrac{1 + \text{cos A}}{1 + \text{cos A}}}

This becomes:

\Rightarrow \sqrt{\dfrac{1 - \text{cos}^2 A}{(1 + \text{cos A})^2}}

Recall the identity:

1 - \text{cos}^2 A = \text{sin}^2 A

Therefore, we have:

\Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{1 + \text{cos A}}

Simplifying further gives:

\Rightarrow \dfrac{\text{sin A}}{\text{1 + cos A}}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}}$ .

Question 2(vi)

Prove the following identities :

\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}

Answer:

We start with the identity:

\text{sec}^2 A - \text{tan}^2 A = 1

Focusing on the left-hand side of the given equation, we have:

\Rightarrow \dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}}

Substituting the identity, it becomes:

\Rightarrow \dfrac{\text{sec}^2 A - \text{tan}^2 A + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}}

This can be rewritten as:

\Rightarrow \dfrac{\text{(sec A - tan A)(sec A + tan A) + (sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}}

Factoring gives:

\Rightarrow \dfrac{\text{(sec A - tan A)[sec A + tan A + sec A - tan A]}}{\text{cosec A (sec A - tan A)}}

Simplifying further:

\Rightarrow \dfrac{\text{2 sec A}}{\text{cosec A}}

Expressing the trigonometric functions in terms of sine and cosine:

\Rightarrow \dfrac{2 \times \dfrac{1}{\text{cos A}}}{\dfrac{1}{\text{sin A}}}

Simplifying the fraction:

\Rightarrow 2\dfrac{\text{sin A}}{\text{cos A}}

Which results in:

\Rightarrow \text{2 tan A}.

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$ .

Question 2(vii)

Prove the following identities :

\dfrac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} = \text{2 cot A}

Answer:

Recall the identity: $\text{cosec}^2 A - \text{cot}^2 A = 1$ .

Let’s focus on the left side of the equation:

\begin{aligned}\Rightarrow \dfrac{(\text{cosec A} - \text{cot A})^2 + 1}{\text{sec A} (\text{cosec A} - \text{cot A})} \\\Rightarrow \dfrac{(\text{cosec A} - \text{cot A})^2 + \text{cosec}^2 A - \text{cot}^2 A}{\text{sec A} (\text{cosec A} - \text{cot A})} \\\Rightarrow \dfrac{(\text{cosec A} - \text{cot A})^2 + (\text{cosec A} - \text{cot A})(\text{cosec A} + \text{cot A})}{\text{sec A} (\text{cosec A} - \text{cot A})} \\\Rightarrow \dfrac{(\text{cosec A} - \text{cot A})(\text{cosec A} - \text{cot A} + \text{cosec A} + \text{cot A})}{\text{sec A} (\text{cosec A} - \text{cot A})} \\\Rightarrow \dfrac{2 \text{cosec A}}{\text{sec A}} \\\Rightarrow \dfrac{\dfrac{2}{\text{sin A}}}{\dfrac{1}{\text{cos A}}} \\\Rightarrow \dfrac{2 \text{cos A}}{\text{sin A}} \\\Rightarrow 2 \text{cot A}.\end{aligned}

Thus, the left-hand side equals the right-hand side, confirming the identity. Hence, proved that ( \dfrac{(\text{cosec A} – \text{cot A})^2 + 1}{\text{sec A} (\text{cosec A} – \text{cot A})} = 2 \text{cot A} ).

Question 2(viii)

Prove the following identities :

$\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0

Answer:

Let’s tackle the left-hand side of the equation:

\Rightarrow \text{cot}^2 A \left(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\times \dfrac{\text{sec A + 1}}{\text{sec A + 1}}\right) + \text{sec}^2 A \left(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\right)

This simplifies to:

\Rightarrow \text{cot}^2 A \left(\dfrac{\text{sec}^2 A - 1}{(1 + \text{sin A})(\text{sec A + 1})}\right) + \text{sec}^2 A \left(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\right)

Recall the identity $\text{sec}^2 A - 1 = \text{tan}^2 A$ .

Substituting this, we get:

\Rightarrow \dfrac{\text{cot}^2 A \times \text{tan}^2 A}{(1 + \text{sin A})(\text{sec A + 1})} + \text{sec}^2 A \left(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\right)

Simplifying further:

\Rightarrow \dfrac{\text{cot}^2 A \times \dfrac{1}{\text{cot}^2 A}}{(1 + \text{sin A})(\text{sec A + 1})} + \text{sec}^2 A \left(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\right)

Which reduces to:

\Rightarrow \dfrac{1}{(1 + \text{sin A})(\text{sec A + 1})} + \text{sec}^2 A \left(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\right)

Combining these terms:

\Rightarrow \dfrac{1 + \text{sec}^2 A(\text{sin A - 1})(1 + \text{sin A})}{(1 + \text{sin A})(\text{sec A + 1})}

Notice the simplification:

\Rightarrow \dfrac{1 - \text{sec}^2 A(1 - \text{sin A})(1 + \text{sin A})}{(1 + \text{sin A})(\text{sec A + 1})}

Recognize that $1 - \text{sin}^2 A = \text{cos}^2 A$ :

\Rightarrow \dfrac{1 - \text{sec}^2 A \times \text{cos}^2 A}{(1 + \text{sin A})(\text{sec A + 1})}

Substitute $\text{sec}^2 A = \dfrac{1}{\text{cos}^2 A}$ :

\Rightarrow \dfrac{1 - \dfrac{1}{\text{cos}^2 A} \times \text{cos}^2 A}{(1 + \text{sin A})(\text{sec A + 1})}

This simplifies to:

\Rightarrow \dfrac{1 - 1}{(1 + \text{sin A})(\text{sec A + 1})}

Finally, we arrive at:

\Rightarrow 0.

Thus, the left-hand side equals the right-hand side, confirming that the identity holds true.

Hence, proved that $\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0.

Question 2(ix)

Prove the following identities :

$\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}$ = 2 cos^2 A – 1

Answer:

Let’s tackle the left-hand side of the given equation:

\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}

Recognize that the denominator resembles a difference of squares, which can be factored as:

\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(\text{cos}^2 A - \text{sin}^2 A)(\text{cos}^2 A + \text{sin}^2 A)}

We know from trigonometric identities that:

\text{cos}^2 A + \text{sin}^2 A = 1

and also,

\text{cos}^2 A = 1 - \text{sin}^2 A.

Substituting these identities, the expression simplifies to:

\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{sin}^2 A - \text{sin}^2 A)}

This reduces to:

\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{2 sin}^2 A)}

Cancel the common term from numerator and denominator:

\Rightarrow \text{1 - 2 sin}^2 A

Now express $\text{sin}^2 A$ in terms of $\text{cos}^2 A$ :

\Rightarrow 1 - 2(\text{1 - cos}^2 A)

Simplifying this gives:

\Rightarrow 1 - 2 + \text{2 cos}^2 A

Finally, we arrive at:

\Rightarrow \text{2 cos}^2 A - 1.

∴ The left-hand side equals the right-hand side, confirming the identity:

Hence, proved that $\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}$ = 2 cos^2 A – 1.

Question 2(x)

Prove the following identities :

sec^4 A (1 – sin^4 A) – 2 tan^2 A = 1

Answer:

Let’s work through the left-hand side of the given identity:

⇒ $\sec^4 A (1 - \sin^4 A) - 2 \tan^2 A$

First, notice that $1 - \sin^4 A$ can be rewritten using the identity $1 - \sin^2 A = \cos^2 A$ :

⇒ $\sec^4 A (1 - \sin^2 A)(1 + \sin^2 A) - 2 \tan^2 A$

Substitute $1 - \sin^2 A$ with $\cos^2 A$ :

⇒ $\sec^4 A \cos^2 A (1 + \sin^2 A) - 2 \tan^2 A$

Now, simplify $\sec^4 A \cos^2 A$ as $\sec^4 A \times \dfrac{1}{\sec^2 A}$ :

⇒ $\sec^2 A (1 + \sin^2 A) - 2 \tan^2 A$

Expanding the expression gives:

⇒ $\sec^2 A + \sec^2 A \sin^2 A - 2 \tan^2 A$

Convert $\sec^2 A \sin^2 A$ into $\dfrac{1}{\cos^2 A} \times \sin^2 A$ :

⇒ $\sec^2 A + \tan^2 A - 2 \tan^2 A$

Simplifying further, we get:

⇒ $\sec^2 A - \tan^2 A$

Using the identity $\sec^2 A - \tan^2 A = 1$ , we find:

⇒ 1.

Thus, the left-hand side equals the right-hand side. Hence, proved that $\sec^4 A (1 - \sin^4 A) - 2 \tan^2 A = 1$ .

Question 2(xi)

Prove the following identities :

(1 + tan A + sec A)(1 + cot A – cosec A) = 2

Answer:

To verify the given identity, let’s focus on the left-hand side (LHS):

\Rightarrow \Big(1 + \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)\Big(1 + \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}\Big)

Simplifying the expressions inside each bracket:

\Rightarrow \Big(\dfrac{\text{cos A} + \text{sin A} + 1}{\text{cos A}}\Big)\Big(\dfrac{\text{sin A} + \text{cos A} - 1}{\text{sin A}}\Big)

Now, multiply these two fractions:

\Rightarrow \dfrac{(\text{sin A} + \text{cos A})^2 - (1)^2}{\text{sin A} \text{cos A}}

Expanding the numerator using the identity ((a + b)^2 = a^2 + 2ab + b^2):

\Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 2 \text{sin A} \text{cos A} - 1}{\text{sin A} \text{cos A}}

Recall the fundamental identity:

\text{sin}^2 A + \text{cos}^2 A = 1

Substitute this into the expression:

\Rightarrow \dfrac{1 - 1 + 2 \text{sin A} \text{cos A}}{\text{sin A} \text{cos A}}

This simplifies to:

\Rightarrow \dfrac{2 \text{sin A} \text{cos A}}{\text{sin A} \text{cos A}}

Cancel the common terms in the numerator and denominator:

\Rightarrow 2.

Thus, the left-hand side equals the right-hand side, confirming the identity. Hence, proved that (1 + \text{tan A} + \text{sec A})(1 + \text{cot A} – \text{cosec A}) = 2.

Question 3

If x = a cos θ and y = b cot θ, show that :

$\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2}$ = 1

Answer:

To prove the given identity, let’s replace the expressions for $x$ and $y$ in the left-hand side of the equation:

\begin{aligned}\Rightarrow \dfrac{a^2}{(a \text{cos } \theta)^2} - \dfrac{b^2}{(b \text{cot } \theta)^2} \\\Rightarrow \dfrac{a^2}{a^2 \text{cos}^2 \theta} - \dfrac{b^2}{b^2 \text{cot}^2 \theta} \\\Rightarrow \dfrac{1}{\text{cos}^2 \theta} - \dfrac{1}{\text{cot}^2 \theta} \\\Rightarrow \dfrac{1}{\text{cos}^2 \theta} - \dfrac{1}{\dfrac{\text{cos}^2 \theta}{\text{sin}^2 \theta}} \\\Rightarrow \dfrac{1}{\text{cos}^2 \theta} - \dfrac{\text{sin}^2 \theta}{\text{cos}^2 \theta} \\\Rightarrow \dfrac{1 - \text{sin}^2 \theta}{\text{cos}^2 \theta} \\\Rightarrow \dfrac{\text{cos}^2 \theta}{\text{cos}^2 \theta} \\\Rightarrow 1.\end{aligned}

Thus, the left-hand side equals the right-hand side, confirming that the identity $\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2} = 1$ holds true.

Question 4

If sec A + tan A = p, show that :

sin A = $\dfrac{p^2 - 1}{p^2 + 1}$

Answer:

Given that $\sec A + \tan A = p$ , we need to express $\sin A$ in terms of $p$ . Let’s start by substituting $p$ into the right-hand side of the equation:

\begin{align*}&\Rightarrow \dfrac{(\sec A + \tan A)^2 - 1}{(\sec A + \tan A)^2 + 1} \\&\Rightarrow \dfrac{\sec^2 A + \tan^2 A + 2 \sec A \tan A - 1}{\sec^2 A + \tan^2 A + 2 \sec A \tan A + 1}\end{align*}

Using the identity $\sec^2 A - 1 = \tan^2 A$ , we can rewrite $\sec^2 A$ as $1 + \tan^2 A$ :

\begin{align*}&\Rightarrow \dfrac{\sec^2 A - 1 + \tan^2 A + 2 \sec A \tan A}{\sec^2 A + \tan^2 A + 1 + 2 \sec A \tan A} \\&\Rightarrow \dfrac{\tan^2 A + \tan^2 A + 2 \sec A \tan A}{\sec^2 A + \sec^2 A + 2 \sec A \tan A} \\&\Rightarrow \dfrac{2 \tan^2 A + 2 \sec A \tan A}{2 \sec^2 A + 2 \sec A \tan A} \\&\Rightarrow \dfrac{2 \tan A (\tan A + \sec A)}{2 \sec A (\sec A + \tan A)} \\&\Rightarrow \dfrac{\tan A}{\sec A}\end{align*}

Now, express $\tan A$ and $\sec A$ in terms of sine and cosine:

\begin{align*}&\Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{1}{\cos A}} \\&\Rightarrow \dfrac{\sin A}{\cos A} \times \cos A \\&\Rightarrow \sin A.\end{align*}

Thus, we have shown that $\sin A = \dfrac{p^2 - 1}{p^2 + 1}$ .

Question 5(i)

If 2 sin A – 1 = 0, show that :

sin 3A = 3 sin A – 4 sin^3 A

Answer:

We start with the equation:

2 \sin A - 1 = 0

Solving for $\sin A$ , we get:

$2 \sin A = 1$
$\sin A = \dfrac{1}{2}$

Recognizing $\sin A = \sin 30^\circ$ , it follows that:

A = 30^\circ

Now, consider the identity:

\sin 3A = 3 \sin A - 4 \sin^3 A

Substitute $A = 30^\circ$ into the left-hand side:

\sin 3A = \sin 3(30^\circ) = \sin 90^\circ = 1

Substituting $A = 30^\circ$ into the right-hand side, we have:

$3 \sin A - 4 \sin^3 A$
$= 3 \sin 30^\circ - 4 \sin^3 30^\circ$
$= 3 \times \dfrac{1}{2} - 4 \times \left( \dfrac{1}{2} \right)^3$
$= \dfrac{3}{2} - 4 \times \dfrac{1}{8}$
$= \dfrac{3}{2} - \dfrac{1}{2}$
$= \dfrac{2}{2}$
$= 1$

Since both sides are equal, $\text{L.H.S.} = \text{R.H.S.}$ , we have shown that the identity holds true:

Hence, proved that $\sin 3A = 3 \sin A - 4 \sin^3 A$ .

Question 5(ii)

If 4 cos^2 A – 3 = 0, show that :

cos 3A = 4 cos^3 A – 3 cos A

Answer:

We start with the equation:

⇒ 4 cos² A – 3 = 0

This implies:

⇒ 4 cos² A = 3

Dividing both sides by 4, we get:

⇒ cos² A = $\dfrac{3}{4}$

Taking the square root of both sides gives:

⇒ cos A = $\sqrt{\dfrac{3}{4}}$

⇒ cos A = $\dfrac{\sqrt{3}}{2}$

Since $\dfrac{\sqrt{3}}{2}$ is the cosine of 30°, we have:

⇒ cos A = cos 30°

⇒ A = 30°

Now, let’s verify the identity.

L.H.S.:

cos 3A = cos 3(30°) = cos 90° = 0

R.H.S.:

4 cos³ A – 3 cos A

Substituting cos A = $\dfrac{\sqrt{3}}{2}$ , we have:

= 4 cos³ 30° – 3 cos 30°

= ( 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 – 3 \times \dfrac{\sqrt{3}}{2} )

= $4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2}$

= $\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2}$

= 0

Since the left-hand side equals the right-hand side, we conclude:

Hence, proved that cos 3A = 4 cos³ A – 3 cos A.

Question 6(i)

Evaluate :

2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)

Answer:

Let’s evaluate the expression step by step:

$2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)$
$= 2\Big(\dfrac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\dfrac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec (90°- 50°)}}{\text{cosec 50°}}\Big)$

Recall the trigonometric identities:

( \text{cot}(90° – \theta) = \text{tan} \theta )
( \text{sec}(90° – \theta) = \text{cosec} \theta )

Applying these identities, we have:

\Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{cosec 50°}}{\text{cosec 50°}}\Big)

Notice that each fraction simplifies to 1:

\Rightarrow 2 \times (1)^2 + (1)^2 - 3 \times 1

Calculate the expression:

\Rightarrow 2 + 1 - 3

\Rightarrow 0.

Thus, the evaluated expression is 0.

Question 6(ii)

sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$

Answer:

To find the value of the expression, we have:

⇒ sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$

Recognizing that sec(90° – θ) is equal to cosec θ, we can rewrite:

⇒ sec (90° – 64°) sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec (90° - 33°)}}$

Applying the identity, it becomes:

⇒ cosec 64° sin 64° + $\dfrac{\text{cosec 33°}}{\text{cosec 33°}}$

Now, notice that cosec 64° is $\dfrac{1}{\text{sin 64°}}$ . Thus:

⇒ $\dfrac{1}{\text{sin 64°}} \times \text{sin 64°} + 1$

The terms simplify to:

⇒ 1 + 1

⇒ 2.

Thus, the result of sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$ is 2.

Question 6(iii)

\dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}

Answer:

To find the value of the given expression:

\Rightarrow \dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}

We can use the identities:

$\sin (90° - \theta) = \cos \theta$
$\cot (90° - \theta) = \tan \theta$

Applying these identities, we rewrite:

\Rightarrow \dfrac{\text{5 sin (90° - 24°)}}{\text{cos 24°}} - \dfrac{\text{2 cot (90° - 5°)}}{\text{tan 5°}}

This simplifies to:

\Rightarrow \dfrac{\text{5 cos 24°}}{\text{cos 24°}} - \dfrac{\text{2 tan 5°}}{\text{tan 5°}}

Notice that $\dfrac{\text{5 cos 24°}}{\text{cos 24°}} = 5$ and $\dfrac{\text{2 tan 5°}}{\text{tan 5°}} = 2$ .

Thus, the expression becomes:

\Rightarrow 5 - 2

Which equals:

\Rightarrow 3.

Question 6(iv)

3 cos 80° cosec 10° + 2 cos 59° cosec 31°

Answer:

Consider the expression:

⇒ 3 \cos (90° – 10°) \cosec 10° + 2 \cos 59° \cosec (90° – 59°)

Utilizing trigonometric identities, we know:

( \cosec (90° – \theta) = \sec \theta ) and ( \cos(90° – \theta) = \sin \theta )

Applying these identities, the expression becomes:

⇒ 3 \sin 10° \cosec 10° + 2 \cos 59° \sec 59°

Breaking it down further:

⇒ 3 \sin 10° \times \dfrac{1}{\text{sin 10°}} + 2 \text{ cos 59°} \times \dfrac{1}{\text{cos 59°}}

Simplifying the terms gives:

⇒ 3 + 2

Thus, the final result is:

⇒ 5.

Therefore, 3 \cos 80° \cosec 10° + 2 \cos 59° \cosec 31° = 5.

Question 7(i)

Prove that :

tan (55° + x) = cot (35° – x)

Answer:

We start with the equation: tan (55° + x) = cot (35° – x).

First, consider the left-hand side:

⇒ tan (55° + x)

This can be rewritten using the identity for tangent of complementary angles:

⇒ tan [90° – (35° – x)]

Recall the trigonometric identity:

\tan (90° - \theta) = \cot \theta

Applying this identity, we get:

⇒ cot (35° – x)

Thus, we have shown that the left-hand side equals the right-hand side.

Therefore, it is proved that tan (55° + x) = cot (35° – x).

Question 7(ii)

sec (70° – θ) = cosec (20° + θ)

Answer:

Consider the equation: sec (70° – θ) = cosec (20° + θ).

Let’s work on the L.H.S:

⇒ sec (70° – θ)

This can be rewritten using the identity:

⇒ sec [90° – (20° + θ)]

According to the trigonometric identity:

sec (90° – θ) = cosec θ

Thus, we have:

⇒ cosec (20° + θ)

Notice that the L.H.S matches the R.H.S.

Hence, proved that sec (70° – θ) = cosec (20° + θ).

Question 7(iii)

sin(28° + A) = cos(62° – A)

Answer:

We start with the equation: sin(28° + A) = cos(62° – A).

Let’s examine the left-hand side:

⇒ sin(28° + A)

This can be rewritten as:

⇒ sin [90° – (62° – A)]

Using the identity:

sin (90° – θ) = cos θ

we can simplify further to:

⇒ cos (62° – A)

Since the left-hand side equals the right-hand side, we have shown that:

Hence, proved that sin(28° + A) = cos(62° – A).

Question 7(iv)

$\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec^2 (90° – A)

Answer:

Using the identity for cosine, we know:

( \cos (90° – A) = \sin A )

Now, let’s work on the left-hand side (L.H.S.) of the equation:

\Rightarrow \dfrac{1}{1 + \sin A} + \dfrac{1}{1 - \sin A}

Combine the fractions:

\Rightarrow \dfrac{1 - \sin A + 1 + \sin A}{(1 + \sin A)(1 - \sin A)}

Simplifying the numerator:

\Rightarrow \dfrac{2}{1 - \sin^2 A}

Recognize the Pythagorean identity:

\Rightarrow \dfrac{2}{\cos^2 A}

This can be rewritten using secant:

\Rightarrow 2 \sec^2 A

Since ( \sec A = \text{cosec}(90° – A) ), we have:

\Rightarrow 2 \text{cosec}^2 (90° - A)

Thus, L.H.S. equals R.H.S., confirming the identity is valid. Hence, proved that ( \dfrac{1}{1 + \cos (90° – A)} + \dfrac{1}{1 – \cos (90° – A)} = 2 \text{cosec}^2 (90° – A) ).

Question 8

If A and B are complementary angles, prove that :

(i) cot B + cos B = sec A cos B (1 + sin B)

(ii) cot A cot B – sin A cos B – cos A sin B = 0

(iii) cosec^2 A + cosec^2 B = cosec^2 A cosec^2 B

(iv) $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$

Answer:

Given that A and B are complementary angles, we have:

A + B = 90°

This implies B = 90° – A and A = 90° – B.

(i) Let’s start by looking at the left-hand side of the equation:

\text{cot B + cos B}

Substitute B with 90° – A:

\Rightarrow \text{cot (90° - A) + cos (90° - A)}

Using trigonometric identities, this becomes:

\Rightarrow \text{tan A + sin A}

Expressing tan A in terms of sine and cosine:

\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \text{sin A}

Combine the terms over a common denominator:

\Rightarrow \dfrac{\text{sin A + sin A cos A}}{\text{cos A}}

Factor out sin A:

\Rightarrow \dfrac{\text{sin A(1 + cos A)}}{\text{cos A}}

This simplifies to:

\Rightarrow \text{sin A sec A (1 + cos A)}

Substituting back for B:

\Rightarrow \text{sin (90° - B) sec A [1 + cos (90° - B)]}

Using the identities:

sin (90° – B) = cos B and cos (90° – B) = sin B

Thus:

\Rightarrow \text{cos B sec A (1 + sin B)}

∴ L.H.S. = R.H.S.

Hence, proved that cot B + cos B = sec A cos B (1 + sin B).

(ii) Now, consider the left-hand side of the equation:

\Rightarrow \text{cot A cot (90° - A) - sin A cos (90° - A) - cos A sin (90° - A)}

Using the identities:

cot (90° – A) = tan A, cos (90° – A) = sin A, and sin (90° – A) = cos A

This becomes:

\Rightarrow \text{cot A tan A - sin A sin A - cos A cos A}

Simplifying further:

\Rightarrow \text{cot A} \times \dfrac{1}{\text{cot A}} - (\text{sin}^2 A + \text{cos}^2 A)

Since sin² A + cos² A = 1:

\Rightarrow 1 - 1

\Rightarrow 0

∴ L.H.S. = R.H.S.

Hence, proved that cot A cot B – sin A cos B – cos A sin B = 0.

(iii) For the next part, consider the left-hand side of the equation:

\Rightarrow \text{cosec}^2 A + \text{cosec}^2 (90° - A)

Using the identity:

cosec (90° – A) = sec A

We have:

\Rightarrow \dfrac{1}{\text{sin}^2 A} + \text{sec}^2 A

Which is:

\Rightarrow \dfrac{1}{\text{sin}^2 A} + \dfrac{1}{\text{cos}^2 A}

Combine the terms:

\Rightarrow \dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A \text{cos}^2 A}

Since sin² A + cos² A = 1:

\Rightarrow \dfrac{1}{\text{sin}^2 A \text{cos}^2 A}

This simplifies to:

\Rightarrow \text{cosec}^2 A \text{sec}^2 A

And further:

\Rightarrow \text{cosec}^2 A \text{sec}^2 (90° - B)

\Rightarrow \text{cosec}^2 A \text{cosec}^2 B

∴ L.H.S. = R.H.S.

Hence, proved that cosec² A + cosec² B = cosec² A cosec² B.

(iv) Finally, let’s simplify the left-hand side:

\Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}}

Substitute using the identities:

\Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos (90° - A) - cos (90° - B)}}{\text{cos (90° - A) + cos (90° - B)}}

Where cos (90° – θ) = sin θ and sin (90° – θ) = cos θ, this becomes:

\Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{sin A - sin B}}{\text{sin A + sin B}}

Combine the fractions:

\Rightarrow \dfrac{(\text{sin A + sin B})^2 + (\text{sin A - sin B})^2}{(\text{sin A - sin B})(\text{sin A + sin B})}

Simplify the numerator:

\Rightarrow \dfrac{\text{sin}^2 A + \text{sin}^2 B + 2 \text{sin A sin B} + \text{sin}^2 A + \text{sin}^2 B - 2 \text{sin A sin B}}{\text{sin}^2 A - \text{sin}^2 B}

This reduces to:

\Rightarrow 2 \dfrac{\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A - \text{sin}^2 B}

Substitute back:

\Rightarrow 2 \dfrac{\text{sin}^2 A + \text{sin}^2 (90° - A)}{\text{sin}^2 A - \text{sin}^2 (90° - A)}

Which is:

\Rightarrow 2 \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}

Since sin² A + cos² A = 1:

\Rightarrow \dfrac{2 \times 1}{\text{sin}^2 A - \text{cos}^2 A}

\Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}

Using the identity:

cos² A = 1 – sin² A

\Rightarrow \dfrac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)}

Simplifying gives:

\Rightarrow \dfrac{2}{\text{sin}^2 A + \text{sin}^2 A - 1}

\Rightarrow \dfrac{2}{\text{2 sin}^2 A - 1}

∴ L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$ .

Question 9(i)

Prove that :

\dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}} = \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}

Answer:

Let’s tackle the left-hand side of the equation:

\Rightarrow \dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}}

Combine the fractions by finding a common denominator:

\Rightarrow \dfrac{\text{sin A + cos A - (sin A - cos A)}}{\text{(sin A - cos A)(sin A + cos A)}}

Simplifying the numerator, we get:

\Rightarrow \dfrac{\text{sin A - sin A + cos A + cos A}}{\text{sin}^2 A - \text{cos}^2 A}

Which simplifies further to:

\Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A - \text{cos}^2 A}

Recall the identity:

\cos^2 A = 1 - \sin^2 A

Substitute this into the denominator:

\Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A - (1 - \text{sin}^2 A)}

Simplify the expression:

\Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A + \text{sin}^2 A - 1}

Which results in:

\Rightarrow \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}

Since the left-hand side equals the right-hand side,

Hence, proved that $\dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}} = \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}$ .

Question 9(ii)

Prove that :

$\dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1$ = cosec A

Answer:

To tackle the left-hand side of the equation, begin with:

\Rightarrow \dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1

This can be rewritten as:

\Rightarrow \dfrac{\text{cot}^2 A - \text{(cosec A - 1)}}{\text{cosec A - 1}}

Simplifying further, we have:

\Rightarrow \dfrac{\text{cot}^2 A + 1 - \text{cosec A}}{\text{cosec A - 1}}

Recall the identity:

\text{cot}^2 A + 1 = \text{cosec}^2 A

Substituting this into our expression gives:

\Rightarrow \dfrac{\text{cosec}^2 A - \text{cosec A}}{\text{cosec A - 1}}

Factor out $\text{cosec A}$ from the numerator:

\Rightarrow \dfrac{\text{cosec A(cosec A - 1)}}{\text{cosec A - 1}}

Cancel the common term $\text{(cosec A - 1)}$ from the numerator and the denominator:

\Rightarrow \text{cosec A}.

Thus, since the left-hand side equals the right-hand side, we have shown that:

Hence, proved that $\dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1$ = cosec A.

Question 9(iii)

Prove that :

$\dfrac{\text{cos A}}{\text{1 + sin A}}$ = sec A – tan A

Answer:

Let’s simplify the right-hand side of the given equation:

⇒ \text{sec A} - \text{tan A}

Expressing sec and tan in terms of sine and cosine:

⇒ \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}

Combine the fractions over a common denominator:

⇒ \dfrac{1 - \text{sin A}}{\text{cos A}}

Multiply numerator and denominator by $1 + \text{sin A}$ :

⇒ \dfrac{1 - \text{sin A}}{\text{cos A}} \times \dfrac{1 + \text{sin A}}{1 + \text{sin A}}

Apply the identity $1 - \text{sin}^2 A = \text{cos}^2 A$ :

⇒ \dfrac{1 - \text{sin}^2 A}{\text{cos A}(1 + \text{sin A})}

Simplify using $1 - \text{sin}^2 A = \text{cos}^2 A$ :

⇒ \dfrac{\text{cos}^2 A}{\text{cos A}(1 + \text{sin A})}

Cancel $\text{cos A}$ from numerator and denominator:

⇒ \dfrac{\text{cos A}}{1 + \text{sin A}}

Thus, the left-hand side equals the right-hand side.

Hence, proved that $\dfrac{\text{cos A}}{1 + \text{sin A}} = \text{sec A} - \text{tan A}$ .

Question 9(iv)

Prove that :

cos A(1 + cot A) + sin A(1 + tan A) = sec A + cosec A

Answer:

Let’s tackle the left-hand side of the equation:

⇒ ( \cos A(1 + \cot A) + \sin A(1 + \tan A) )

Expanding each term, we have:

⇒ $\cos A + \cos A \cot A + \sin A + \sin A \tan A$

Substituting the identities for $\cot A$ and $\tan A$ , we get:

⇒ $\cos A + \cos A \times \dfrac{\cos A}{\sin A} + \sin A + \sin A \times \dfrac{\sin A}{\cos A}$

This simplifies to:

⇒ $\cos A + \dfrac{\sin^2 A}{\cos A} + \sin A + \dfrac{\cos^2 A}{\sin A}$

Combining the fractions, we have:

⇒ $\dfrac{\cos^2 A + \sin^2 A}{\cos A} + \dfrac{\sin^2 A + \cos^2 A}{\sin A}$

Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ , this becomes:

⇒ $\dfrac{1}{\cos A} + \dfrac{1}{\sin A}$

Finally, this simplifies to:

⇒ $\sec A + \text{cosec } A$ .

Thus, the left-hand side equals the right-hand side.

Hence, proved that ( \cos A(1 + \cot A) + \sin A(1 + \tan A) = \sec A + \text{cosec } A ).

Question 9(v)

$\sqrt{\text{sec}^2 A + \text{cosec}^2 A}$ = tan A + cot A

Answer:

Let’s tackle the left-hand side (L.H.S.) of the equation first:

\Rightarrow \sqrt{\left(\frac{1}{\cos A}\right)^2 + \left(\frac{1}{\sin A}\right)^2}

This simplifies to:

\Rightarrow \sqrt{\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}}

Combine the fractions into a single expression:

\Rightarrow \sqrt{\frac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A}}

Using the identity $\sin^2 A + \cos^2 A = 1$ , we get:

\Rightarrow \sqrt{\frac{1}{\cos^2 A \sin^2 A}}

This further simplifies to:

\Rightarrow \frac{1}{\sin A \cos A}

Now, let’s consider the right-hand side (R.H.S.) of the equation:

\Rightarrow \tan A + \cot A

This is expressed as:

\Rightarrow \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}

Combining these fractions, we have:

\Rightarrow \frac{\sin^2 A + \cos^2 A}{\cos A \sin A}

Again, using the identity $\sin^2 A + \cos^2 A = 1$ :

\Rightarrow \frac{1}{\sin A \cos A}

Notice that the L.H.S. equals the R.H.S.

Hence, proved that $\sqrt{\text{sec}^2 A + \text{cosec}^2 A}$ = tan A + cot A.

Question 9(vi)

(sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A

Answer:

Let’s explore the left-hand side of the equation:

\Rightarrow \text{sin A sec A + sin A cosec A + cos A sec A + cos A cosec A}

Breaking it down further:

\Rightarrow \text{sin A} \times \dfrac{1}{\text{cos A}} + \text{sin A} \times \dfrac{1}{\text{sin A}} + \text{cos A} \times \dfrac{1}{\text{cos A}} + \text{cos A} \times \dfrac{1}{\text{sin A}}

Simplifying each term:

\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + 1 + \dfrac{\text{cos A}}{\text{sin A}}

Notice that $1 + 1 = 2$ , so:

\Rightarrow 2 + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}

Since $\text{sin}^2 A + \text{cos}^2 A = 1$ , it becomes:

\Rightarrow 2 + \dfrac{1}{\text{sin A cos A}}

Thus, we have:

$\Rightarrow 2 + \text{cosec A sec A}$ .

This confirms that the left-hand side equals the right-hand side.

Hence, proved that (sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A.

Question 9(vii)

(tan A + cot A)(cosec A – sin A)(sec A – cos A) = 1

Answer:

To solve the left-hand side of the equation, we start with:

\Rightarrow (\text{tan A + cot A})(\text{cosec A - sin A})(\text{sec A - cos A})

First, rewrite each trigonometric function in terms of sine and cosine:

\Rightarrow \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big)\Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big)

Combine the fractions:

\Rightarrow \Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}\Big)\Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big)\Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big)

Recall the identities:

$\text{sin}^2 A + \text{cos}^2 A = 1$
$1 - \text{sin}^2 A = \text{cos}^2 A$
$1 - \text{cos}^2 A = \text{sin}^2 A$

Substitute these into the equation:

\Rightarrow \dfrac{1}{\text{sin A cos A}} \times \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}}

Simplify the expression:

\Rightarrow \dfrac{\text{sin}^2 A \text{cos}^2 A}{\text{sin}^2 A \text{cos}^2 A}

This simplifies to 1.

Since L.H.S. = R.H.S.,

Hence, proved that (tan A + cot A)(cosec A – sin A)(sec A – cos A) = 1.

Question 9(viii)

cot^2 A – cot^2 B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec^2 A – cosec^2 B

Answer:

To solve this, let’s start with the given expression:

\Rightarrow \text{cot}^2 A - \text{cot}^2 B

This can be rewritten using the identity for cotangent:

\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \dfrac{\text{cos}^2 B}{\text{sin}^2 B}

To combine these fractions, we find a common denominator:

\Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 B - \text{cos}^2 B \text{ sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B}

Here, the identity $\sin^2 \theta = 1 - \cos^2 \theta$ is useful. Applying it:

\Rightarrow \dfrac{\text{cos}^2 A (1 - \text{ cos}^2 B) - \text{cos}^2 B (1 - \text{ cos}^2 A)}{\text{sin}^2 A \text{ sin}^2 B}

Expanding the terms gives:

\Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 A \text{cos}^2 B - \text{cos}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B}

Simplifying further:

\Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B}

Notice that this can be rewritten using the sine identity:

\Rightarrow \dfrac{1 - \text{sin}^2 A - (1 - \text{sin}^2 B)}{\text{sin}^2 A \text{ sin}^2 B}

This simplifies to:

\Rightarrow \dfrac{1 - 1 - \text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B}

Further simplification yields:

\Rightarrow \dfrac{-\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B}

Breaking this into two separate fractions:

\Rightarrow -\dfrac{\text{sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} + \dfrac{\text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B}

Simplifying each fraction gives:

\Rightarrow -\dfrac{1}{\text{sin}^2 B} + \dfrac{1}{\text{sin}^2 A}

This can be expressed in terms of cosecant:

\Rightarrow -\text{cosec}^2 B + \text{cosec}^2 A

Thus, we have:

\Rightarrow \text{cosec}^2 A - \text{cosec}^2 B.

Hence, proved that cot^2 A – cot^2 B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec^2 A – cosec^2 B.

Question 9(ix)

Prove that :

\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}

Answer:

Let’s start by simplifying the left-hand side (L.H.S.) of the equation:

First, express cotangent and secant in terms of tangent:

\Rightarrow \dfrac{\dfrac{1}{\text{tan A}} - 1}{2 - (1 + \text{tan}^2 A)}

This simplifies to:

\Rightarrow \dfrac{\dfrac{1 - \text{tan A}}{\text{tan A}}}{2 - 1 - \text{tan}^2 A}

Now, notice the denominator:

\Rightarrow \dfrac{1 - \text{tan A}}{\text{tan A(1 - tan}^2 A)}

Recognize that $1 - \text{tan}^2 A$ can be factored as ((1 – \text{tan A})(1 + \text{tan A})):

\Rightarrow \dfrac{1 - \text{tan A}}{\text{tan A(1 - tan A)(1 + tan A)}}

Cancel ((1 – \text{tan A})) from the numerator and denominator:

\Rightarrow \dfrac{1}{\text{tan A(1 + tan A)}}

Rewriting $\text{tan A}$ using its reciprocal relationship with $\text{cot A}$ , we get:

\Rightarrow \dfrac{1}{\dfrac{1}{\text{cot A}} \text{(1 + tan A)}}

This simplifies to:

\Rightarrow \dfrac{\text{cot A}}{\text{1 + tan A}}

Since the left-hand side equals the right-hand side, we have shown that:

Hence, proved that $\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}$ .

Question 10

If 4 cos^2 A – 3 = 0 and 0° ≤ A ≤ 90°; then prove that :

(i) sin 3A = 3 sin A – 4 sin^3 A

(ii) cos 3A = 4 cos^3 A – 3 cos A

Answer:

We start with the given equation:

4 \cos^2 A - 3 = 0

This simplifies to:

4 \cos^2 A = 3

Dividing both sides by 4 gives:

\cos^2 A = \dfrac{3}{4}

Taking the square root, we find:

\cos A = \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}

Since $\cos A = \cos 30^\circ$ , it follows that:

A = 30^\circ

(i) Proving $\sin 3A = 3 \sin A - 4 \sin^3 A$ :

Let’s evaluate the left side:

\sin 3A = \sin 3(30^\circ) = \sin 90^\circ = 1

Now, consider the right side:

3 \sin A - 4 \sin^3 A

Substituting $\sin 30^\circ = \dfrac{1}{2}$ :

3 \times \dfrac{1}{2} - 4 \times \left(\dfrac{1}{2}\right)^3

This simplifies to:

\dfrac{3}{2} - 4 \times \dfrac{1}{8}

Further simplification gives:

\dfrac{3}{2} - \dfrac{1}{2} = \dfrac{2}{2} = 1

Since the left side equals the right side, we have shown:

Therefore, $\sin 3A = 3 \sin A - 4 \sin^3 A$ is proven.

(ii) Proving $\cos 3A = 4 \cos^3 A - 3 \cos A$ :

Evaluate the left side:

\cos 3A = \cos 3(30^\circ) = \cos 90^\circ = 0

Now, consider the right side:

4 \cos^3 A - 3 \cos A

Substituting $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ :

4 \times \left(\dfrac{\sqrt{3}}{2}\right)^3 - 3 \times \dfrac{\sqrt{3}}{2}

This simplifies to:

4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2}

Further simplification gives:

\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} = 0

Since the left side equals the right side, we have shown:

Therefore, $\cos 3A = 4 \cos^3 A - 3 \cos A$ is proven.

Question 11

Find A, if 0° ≤ A ≤ 90° and :

(i) 2 cos^2 A – 1 = 0

(ii) sin 3A – 1 = 0

(iii) 4 sin^2 A – 3 = 0

(iv) cos^2 A – cos A = 0

(v) 2 cos^2 A + cos A – 1 = 0

Answer:

(i) Let’s work through this:

⇒ Start with the equation: 2 \cos^2 A – 1 = 0

⇒ Rearrange to find: 2 \cos^2 A = 1

⇒ Dividing both sides by 2 gives: \cos^2 A = \dfrac{1}{2}

⇒ Taking the square root, we have: \cos A = \sqrt{\dfrac{1}{2}}

⇒ Simplifying, \cos A = \dfrac{1}{\sqrt{2}}

⇒ Recognizing the angle, \cos A = \cos 45°

⇒ Therefore, A = 45°.

Hence, A = 45°.

(ii) Consider the equation:

⇒ sin 3A – 1 = 0

⇒ This simplifies to: sin 3A = 1

⇒ We know sin 3A = sin 90°

⇒ Thus, 3A = 90°

⇒ Solving for A gives: A = 30°.

Hence, A = 30°.

(iii) Now, let’s solve:

⇒ 4 \sin^2 A – 3 = 0

⇒ Rearrange to get: 4 \sin^2 A = 3

⇒ Dividing by 4, we find: \sin^2 A = \dfrac{3}{4}

⇒ Taking the square root gives: \sin A = \sqrt{\dfrac{3}{4}}

⇒ Simplifying, \sin A = \dfrac{\sqrt{3}}{2}

⇒ Recognizing the angle, \sin A = \sin 60°

⇒ Therefore, A = 60°.

Hence, A = 60°.

(iv) Let’s solve this equation:

⇒ \cos^2 A – \cos A = 0

⇒ Factoring, we have: \cos A(\cos A – 1) = 0

⇒ So, \cos A = 0 or \cos A – 1 = 0

⇒ This means \cos A = 0 or \cos A = 1

⇒ Recognizing the angles, \cos A = \cos 90° or \cos A = \cos 0°

⇒ Therefore, A = 90° or A = 0°.

Hence, A = 0° or 90°.

(v) Now, solve:

⇒ 2 \cos^2 A + \cos A – 1 = 0

⇒ Rearrange: 2 \cos^2 A + 2 \cos A – \cos A – 1 = 0

⇒ Factoring, we get: 2 \cos A(\cos A + 1) – 1(\cos A + 1) = 0

⇒ Further simplifying, (2 \cos A – 1)(\cos A + 1) = 0

⇒ This implies 2 \cos A = 1 or \cos A = -1

⇒ Solving gives \cos A = \dfrac{1}{2} or \cos A = -1

Since \cos A cannot be negative for 0° ≤ A ≤ 90°,

∴ \cos A = \dfrac{1}{2}

⇒ Recognizing the angle, \cos A = \cos 60°

⇒ Therefore, A = 60°

Hence, A = 60°.

Question 12

If 0° < A < 90°; find A if :

(i) $\dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 4$

(ii) $\dfrac{\text{sin A}}{\text{sec A - 1}} + \dfrac{\text{sin A}}{\text{sec A + 1}}$ = 2

Answer:

(i) Let’s tackle the left-hand side of the equation:

\dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 4

Combine the fractions:

\dfrac{\text{cos A(1 + sin A) + cos A(1 - sin A)}}{\text{(1 + sin A)(1 - sin A)}}

This simplifies to:

\dfrac{\text{cos A + cos A sin A + cos A - cos A sin A}}{\text{1 - sin}^2 A}

Recall that $1 - \sin^2 A = \cos^2 A$ . Substituting gives:

\dfrac{\text{2 cos A}}{\text{cos}^2 A}

Simplifying further:

\dfrac{2}{\text{cos A}} = 2\text{ sec A}

We know the right-hand side is 4:

∴ $2 \text{ sec A} = 4$

⇒ $\text{sec A} = 2$

This implies:

⇒ $\text{sec A} = \text{sec } 60^\circ$

⇒ $A = 60^\circ$ .

Hence, A = 60°.

(ii) Now, for the left-hand side of this equation:

\dfrac{\text{sin A}}{\text{sec A - 1}} + \dfrac{\text{sin A}}{\text{sec A + 1}} = 2

Combine these fractions:

\dfrac{\text{sin A(sec A + 1) + sin A(sec A - 1)}}{\text{(sec A - 1)(sec A + 1)}}

This simplifies to:

\dfrac{\text{sin A sec A + sin A + sin A sec A - sin A}}{\text{sec}^2 A - 1}

Recall that $\text{sec}^2 A - 1 = \text{tan}^2 A$ . Substituting gives:

\dfrac{\text{2 sin A sec A}}{\text{tan}^2 A}

Express $\text{sin A sec A}$ as $\text{2 sin A} \times \dfrac{1}{\text{cos A}}$ :

\dfrac{\text{2 tan A}}{\text{tan}^2 A}

Simplifying further:

\dfrac{2}{\text{tan A}} = \text{2 cot A}

We know the right-hand side is 2:

∴ $2 \text{ cot A} = 2$

⇒ $\text{cot A} = 1$

This implies:

⇒ $\text{cot A} = \text{cot } 45^\circ$

⇒ $A = 45^\circ$ .

Hence, A = 45°.

Question 13

Prove that :

(cosec A – sin A)(sec A – cos A) sec^2 A = tan A

Answer:

Let’s tackle the left-hand side of the equation:

\Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big)\text{sec}^2 \text{A}

Rewrite each part using identities:

\Rightarrow \Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big) \times \text{sec}^2 A

Recall the Pythagorean identities:

1 – sin² A = cos² A and 1 – cos² A = sin² A

Substitute these into the expression:

\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos}^2 A}

This simplifies to:

\Rightarrow \dfrac{\text{sin A}}{\text{cos A}}

Which is the definition of:

\Rightarrow \text{tan A}.

Thus, the left-hand side equals the right-hand side, confirming the identity is correct.

Hence, proved that (cosec A – sin A)(sec A – cos A) sec^2 A = tan A

Question 14

Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Answer:

Consider the expression on the left-hand side:

\Rightarrow \text{(sin θ + cos θ)(tan θ + cot θ)}

We can rewrite $\text{tan θ}$ and $\text{cot θ}$ using their definitions:

\Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{\text{cos θ}}{\text{sin θ}}\Big)

Next, combine the fractions within the parentheses:

\Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin}^2 θ + \text{cos}^2 θ}{\text{cos θ sin θ}}\Big)

Recall the Pythagorean identity $\text{sin}^2 θ + \text{cos}^2 θ = 1$ :

\Rightarrow \text{(sin θ + cos θ)} \times \dfrac{1}{\text{cos θ sin θ}}

Distribute the terms:

\Rightarrow \dfrac{\text{sin θ}}{\text{cos θ sin θ}} + \dfrac{\text{cos θ}}{\text{cos θ sin θ}}

Simplify each term:

\Rightarrow \dfrac{1}{\text{cos θ}} + \dfrac{1}{\text{sin θ}}

This gives us the right-hand side:

\Rightarrow \text{sec θ + cosec θ}

Thus, we have shown that $(\text{sin θ + cos θ})(\text{tan θ + cot θ}) = \text{sec θ + cosec θ}$ .

Question 15

Evaluate without using trigonometric tables,

sin^2 28° + sin^2 62° + tan^2 38° – cot^2 52° + $\dfrac{1}{4} \text{sec}^2 \space 30°$

Answer:

To evaluate the expression, begin with:

⇒ $\sin^2 28° + \sin^2 62° + \tan^2 38° - \cot^2 52° + \dfrac{1}{4} \text{sec}^2 \space 30°$

Notice that $\sin^2 62°$ can be rewritten using the identity $\sin(90° - \theta) = \cos \theta$ . Similarly, $\tan^2 38°$ can be expressed using $\tan(90° - \theta) = \cot \theta$ :

⇒ $\sin^2 28° + \sin^2 (90° - 28°) + \tan^2 (90° - 52°) - \cot^2 52° + \dfrac{1}{4} \times \Big(\dfrac{2}{\sqrt{3}}\Big)^2$

This simplifies to:

⇒ $\sin^2 28° + \cos^2 28° + \cot^2 52° - \cot^2 52° + \dfrac{1}{4} \times \dfrac{4}{3}$

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ , we have:

⇒ $1 + \dfrac{1}{3}$

Combining these terms gives:

⇒ $1\dfrac{1}{3}$ .

Thus, the value of $\sin^2 28° + \sin^2 62° + \tan^2 38° - \cot^2 52° + \dfrac{1}{4} \text{sec}^2 30°$ is $1\dfrac{1}{3}$ .

Frequently Asked Questions

The three fundamental Pythagorean identities are: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ. Most proofs in this chapter rely on manipulating expressions to use one of these core identities.

To prove an identity, you typically start with the more complex side of the equation, usually the Left-Hand Side (LHS). You then use algebraic manipulation and fundamental identities to simplify it step-by-step until it becomes identical to the other side, the Right-Hand Side (RHS).

A trigonometric identity is true for all possible values of the variable angle, whereas a trigonometric equation is only true for specific values of the variable. For example, sin²θ + cos²θ = 1 is an identity, but sin θ = 1/2 is an equation.

Yes, these solutions are prepared by subject matter experts and follow the exact step-by-step method required by the ICSE board. Practising these problems will prepare you thoroughly for the types of questions on trigonometrical identities asked in board examinations.

ICSE Board

Exercise 21(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Exercise 21(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 2(vii)

Question 2(viii)

Question 2(ix)

Question 3

Question 4

Question 5

Question 6

Exercise 21(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 3

Question 4(i)

Question 4(ii)

Question 5

Question 6(i)

Question 6(ii)

Question 6(iii)

Question 6(iv)

Question 6(v)

Question 6(vi)

Question 6(vii)

Question 7

Question 8(i)

Question 8(ii)