Trigonometrical Identities Class 10 ICSE — Bansal Chapter 21
The trigonometrical identities Class 10 ICSE chapter in the Bansal textbook (Chapter 21) builds the ratio work from Class IX into a complete toolkit: six ratios, relations among them, identity proofs, complementary angles, and the trigonometrical tables. This page explains each section in book order, with formulas and worked patterns. The chapter PDF is downloadable from this page.
Download Trigonometrical Identities Class 10 ICSE Mathematics Chapter 21 PDF
Download the Trigonometrical Identities Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 16 pages, 6 sections, 87 questions.
What is inside Chapter 21
16 pages · 6 sections · 11 worked examples · 87 exercise questions
| Section | Title |
|---|---|
| 21.1 | Trigonometry |
| 21.2 | Trigonometrical Ratios |
| 21.3 | Relations Between Different Trigonometrical Ratios |
| 21.4 | Trigonometric Identities |
| 21.5 | Trigonometrical Ratios of Complementary Angles |
| 21.6 | Using the Trigonometrical Tables |
- EXERCISE 21(A): 48 questions
- EXERCISE 21(B): 7 questions
- EXERCISE 21(C): 11 questions
- EXERCISE 21(D): 21 questions
21.1 Trigonometry
Trigonometry means the science of measuring triangles, and that is the whole aim of this chapter. Every later section measures a right triangle from one of its acute angles, then connects those measurements.
The six ratios of an acute angle come first (21.2), followed by the reciprocal, quotient and square relations among them (21.3). Those relations build the identities (21.4), then complementary angles (21.5), then the printed tables for any acute angle (21.6).
The book expects the ratio foundations from Class IX. If those feel distant, 21.2 restates them for a right-angled triangle before any proof work begins. For syllabus and specimen-paper updates, check the CISCE website.
21.2 Trigonometrical Ratios
For an acute angle A of a right-angled triangle, six ratios compare pairs of sides. With triangle ABC right-angled at B, BC is the perpendicular opposite A, AB is the base next to A, and AC is the hypotenuse.
- \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC}
- \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC}
- \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB}
- \cot A = \frac{\text{Base}}{\text{Perpendicular}} = \frac{AB}{BC}
- \sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{AC}{AB}
- \operatorname{cosec} A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{BC}
Two properties make the ratios useful. Every ratio is a real number with no unit. For a fixed angle, the ratio is the same in any size of triangle, because the triangles share the same angles and are similar.
Worked pattern: take a 3-4-5 triangle and a 6-8-10 triangle with the same acute angle A.
\sin A = \frac{3}{5} = \frac{6}{10}, \qquad \cos A = \frac{4}{5} = \frac{8}{10}
Point: the units cancel, so each ratio is the same pure number in both triangles.
The usual early mistake is swapping perpendicular and base when the triangle is rotated. Fix the position of angle A first, then name the sides around it.
21.3 Relations Between Different Trigonometrical Ratios
The six ratios fall into three families, and every identity proof in 21.4 draws on them. Learn the families in the order the book gives.
Reciprocal relations: sine pairs with cosecant, cosine with secant, tangent with cotangent. Each pair multiplies to 1.
- \sin A = \frac{1}{\operatorname{cosec} A}, \operatorname{cosec} A = \frac{1}{\sin A}
- \cos A = \frac{1}{\sec A}, \sec A = \frac{1}{\cos A}
- \tan A = \frac{1}{\cot A}, \cot A = \frac{1}{\tan A}
Quotient relations: dividing sine by cosine cancels the hypotenuse and leaves tangent.
\tan A = \frac{\sin A}{\cos A}, \qquad \cot A = \frac{\cos A}{\sin A}Square relations: these come from Pythagoras. In the right triangle, AB^2 + BC^2 = AC^2, and dividing through by AC^2, AB^2 or BC^2 gives the three forms. That is why they hold for every value of A.
\sin^2 A + \cos^2 A = 1, \qquad 1 + \tan^2 A = \sec^2 A, \qquad 1 + \cot^2 A = \operatorname{cosec}^2 AThe rearranged forms matter for substitution. The table below collects them.
| Square relation | Rearranged forms |
|---|---|
| \sin^2 A + \cos^2 A = 1 | \sin^2 A = 1 - \cos^2 A, \cos^2 A = 1 - \sin^2 A |
| 1 + \tan^2 A = \sec^2 A | \sec^2 A - \tan^2 A = 1, \sec^2 A - 1 = \tan^2 A |
| 1 + \cot^2 A = \operatorname{cosec}^2 A | \operatorname{cosec}^2 A - \cot^2 A = 1, \operatorname{cosec}^2 A - 1 = \cot^2 A |
Check the first relation at 30°. Since \sin 30^\circ = \frac{1}{2} and \cos 30^\circ = \frac{\sqrt{3}}{2}, the squares add to \frac{1}{4} + \frac{3}{4} = 1.
One notation slip to avoid: \sin^2 A means the square of \sin A, written (\sin A)^2, not the sine of A^2.
21.4 Trigonometric Identities
An equation in the trigonometrical ratios of an angle A that is true for all values of A is a trigonometrical identity. The reciprocal, quotient and square relations of 21.3 are all identities. Identity proofs in Class 10 ICSE Maths rely on these three families, so 21.3 must come first.
The proof method: start with one side, usually the more complicated one, and reduce it step by step to the other side. Start with the harder side because it has more to simplify, which makes the next move easier to see. If both sides are complicated, reduce both separately to the same result.
Original worked proof: show that \sin A \sec A \cot A = 1.
Step 1: rewrite \sec A and \cot A in sine and cosine.
\sin A \sec A \cot A = \sin A \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A}
Step 2: cancel \cos A and \sin A.
\sin A \sec A \cot A = 1 = \text{R.H.S.}
Hence proved.
The book’s solved examples cycle through four rewrite moves:
- Rewrite everything in sine and cosine.
- Apply a square relation, usually \sin^2 A + \cos^2 A = 1.
- Factor and cancel common factors.
- Multiply numerator and denominator by a conjugate such as 1 + \cos A.
Some questions carry year tags from past board papers. The tags mark patterns the papers repeat, so they are worth learning early.
| Identity pattern printed in the book | Year tags |
|---|---|
| \frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2\operatorname{cosec} A | 2004, 2009 |
| \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} | 2000, 2013 |
| \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \cos A + \sin A | 2015 |
21.5 Trigonometrical Ratios of Complementary Angles
Two acute angles are complementary when they add to 90°. For any acute angle A, the six complementary relations are:
- \sin (90^\circ - A) = \cos A
- \cos (90^\circ - A) = \sin A
- \tan (90^\circ - A) = \cot A
- \cot (90^\circ - A) = \tan A
- \sec (90^\circ - A) = \operatorname{cosec} A
- \operatorname{cosec} (90^\circ - A) = \sec A
The pattern is clear in a right triangle. In triangle ABC with B = 90^\circ, the angles A and C are complementary. The side opposite A is the side adjacent to C, so \sin A = \cos C.
Worked pattern 1: evaluate \tan 35^\circ \tan 55^\circ.
\tan 55^\circ = \tan (90^\circ - 35^\circ) = \cot 35^\circ
So \tan 35^\circ \tan 55^\circ = \tan 35^\circ \cot 35^\circ = 1.
Worked pattern 2: find x if \sin x = \cos 40^\circ.
\cos 40^\circ = \sin (90^\circ - 40^\circ) = \sin 50^\circ
Final answer: x = 50^\circ.
For a triangle ABC, the relations extend to half-angles. Since A + B + C = 180^\circ, we get \frac{A+B}{2} = 90^\circ - \frac{C}{2}, and therefore \sin \frac{A+B}{2} = \cos \frac{C}{2}. The book’s exercise uses this form.
21.6 Using the Trigonometrical Tables
The trigonometrical tables list natural sines, cosines and tangents to four decimal places. Each table has three parts: the left column with degrees from 0° to 89°, ten columns for the minutes 0′, 6′, 12′, 18′, 24′, 30′, 36′, 42′, 48′ and 54′, and five difference columns for 1′ to 5′.
One degree divides into sixty minutes, so 1^\circ = 60'. An angle like 42° 35′ is not printed directly. Split it into a listed row plus minutes, then apply the difference.
Worked pattern 1: find \sin 42^\circ 35'.
Step 1: write 42^\circ 35' = 42^\circ 30' + 5'. The table gives \sin 42^\circ 30' = 0.6756.
Step 2: the 5′ difference, about 0.0011, is added for sine.
\sin 42^\circ 35' = 0.6756 + 0.0011 = 0.6767
Worked pattern 2 (the cosine flip): find \cos 61^\circ 17'.
Step 1: write 61^\circ 17' = 61^\circ 12' + 5'. The table gives \cos 61^\circ 12' = 0.4818.
Step 2: the difference is subtracted for cosine, because cosine falls as the angle rises.
\cos 61^\circ 17' = 0.4818 - 0.0013 = 0.4805
Reverse lookup: find \theta if \sin \theta = 0.6244.
Step 1: locate the tabular value just below 0.6244; \sin 38^\circ 36' = 0.6239.
Step 2: the residue is 0.6244 - 0.6239 = 0.0005, which matches a 2′ difference.
Final answer: \theta = 38^\circ 36' + 2' = 38^\circ 38'.
Common mistakes to avoid:
- Adding the minute difference for cosine. It is subtracted, not added.
- Reading 1° as 100 minutes. 1^\circ = 60'.
- Confusing \sin^2 A = (\sin A)^2 with \sin (A^2) when checking table values against standard angles.
The reverse cosine rule is the mirror image: a larger cosine value belongs to a smaller angle, so the minutes come off. These table skills carry directly into the Heights and Distances chapter, where ratios meet real objects.
What to practise from Trigonometrical Identities Class 10 ICSE
The chapter ends with four exercise blocks, 87 questions in total. Work through them in the order that matches your weak spots.
- EXERCISE 21(A) — 48 questions on proving identities.
- EXERCISE 21(B) — 7 questions mixing fractional proofs and if-then statements with parameters such as m and n.
- EXERCISE 21(C) — 11 questions on complementary angles, including triangle-based proofs.
- EXERCISE 21(D) — 21 questions on table use, complementary angle evaluation, and harder identities.
A practical order is 21(C) first as the shortest block, then 21(A) in batches, then 21(B), then 21(D). Several questions print the year of a board paper in which they appeared, spanning 2000 to 2015, as a signal of repeated question types.
Trigonometrical Identities Class 10 ICSE frequently asked questions
How do I prove trigonometrical identities class 10 ICSE problems without getting stuck?
Rewrite every ratio in sine and cosine first. Then apply a square relation such as \sin^2 A + \cos^2 A = 1, then factor and cancel. Start from the more complicated side and work toward the simpler side.
Which side should I start with when proving a trigonometrical identity?
In general, start with the more complicated side and reduce it to the other side. If both sides look complicated, reduce both separately to the same result.
Why do I subtract the minute difference for cosine but add it for sine and tangent?
As an acute angle grows, sine and tangent increase, so extra minutes push their values up. Cosine decreases as the angle grows, so the same minutes pull its value down.
When do I use complementary angle ratios like sin(90° − A) = cos A?
Use them when an angle is above 45°, or when an expression mixes two angles that add to 90°. They rewrite an angle as its complement and evaluate products like \tan 35^\circ \tan 55^\circ without tables.
How do I find an acute angle from a sine, cosine or tangent value using the tables?
Find the tabular value just below your given sine or tangent, note its degrees and minutes, then add the minute difference that covers the residue. For cosine, subtract the minutes, because a larger cosine belongs to a smaller angle.
Reference: ICSE Class 10 Mathematics textbook material, chapter 21: Trigonometrical Identities — CISCE-aligned.
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