Complementary Angles Class 9 ICSE — Chapter 25
Complementary angles class 9 ICSE is the chapter that links the six trigonometric ratios through one idea: the two acute angles of a right triangle always add to 90°. In Chapter 25 of your Class 9 Mathematics book, you will see why the sine of an angle equals the cosine of its complement. That single rule lets you evaluate and prove trigonometric expressions without a calculator.
Download Complementary Angles Class 9 ICSE Mathematics Chapter 25 PDF
Download the Complementary Angles Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 6 pages, 3 sections, 8 questions.
What is inside Chapter 25
6 pages · 3 sections · 5 worked examples · 8 exercise questions
| Section | Title |
|---|---|
| 25.1 | INTRODUCTION |
| 25.2 | CONCEPT OF TRIGONOMETIC RATIOS OF COMPLEMENTARY ANGLES |
| 25.3 | COMPLEMENTARY ANGLES FOR SINE (sin) AND COSINE (cos) |
- EXERCISE 25: 8 questions
25.1 INTRODUCTION
Before this chapter, you studied the trigonometric ratios of standard angles: 0°, 30°, 45°, 60° and 90°. You also practised evaluating expressions that combine those ratios. That earlier work gave you a table of exact values to draw on. The foundation came from right triangles and their side ratios.
This chapter moves beyond the standard angles. It deals with any acute angle and its complement — the angle that adds to it to make 90°. You will see why the ratios of complementary angles are linked. Those links let you evaluate expressions faster and prove identities cleanly.
The work builds on the right-triangle methods from the previous chapter, solution of right triangles. Once you can name the perpendicular, base and hypotenuse of a right triangle, the complementary angle identities follow from the same side ratios. Review that chapter if the ratio definitions feel shaky.
25.2 CONCEPT OF TRIGONOMETIC RATIOS OF COMPLEMENTARY ANGLES
Two acute angles are complementary when their sum is exactly 90°. The name comes from one angle ‘completing’ a right angle. If you know one acute angle, you can always find the other by subtracting from 90°.
The chapter gives three patterns to fix the idea. 30° and 60° are complementary because 30^\circ + 60^\circ = 90^\circ. So are 70° and 20°. In general, x^\circ and (90 - x)^\circ are complementary, since their sum is 90^\circ.
For an angle \theta^\circ, the complement is (90 - \theta)^\circ. For example, the complement of 51^\circ is 90^\circ - 51^\circ = 39^\circ. You will use this subtraction constantly in the next section, so it is worth practising until it is instant.
This definition is the base for every identity in the chapter. You will also use it across the Class 9 maths chapters as the year progresses.
25.3 COMPLEMENTARY ANGLES FOR SINE (sin) AND COSINE (cos)
Take triangle ABC with \angle B = 90^\circ and \angle ACB = \theta. The third angle \angle CAB is 90^\circ - \theta. Compare the ratios from each corner of the same triangle.
At \theta, the perpendicular is AB, so \sin \theta = \frac{AB}{AC}. At the complement, that same side AB is the base, so \cos(90^\circ - \theta) = \frac{AB}{AC}. The fractions are identical. This is why the sine of an angle equals the cosine of its complement: both are the same side ratio, named from different corners.
Sine and cosine
The same comparison with the other side gives the reverse. \cos \theta = \frac{BC}{AC} and \sin(90^\circ - \theta) = \frac{BC}{AC}. So \sin(90^\circ - \theta) = \cos \theta.
Tangent and cotangent
Tangent is perpendicular over base. \tan \theta = \frac{AB}{BC}, and \cot(90^\circ - \theta) = \frac{AB}{BC}. Hence \tan \theta = \cot(90^\circ - \theta). The reverse \cot \theta = \tan(90^\circ - \theta) also holds.
Secant and cosecant
Secant is hypotenuse over base. \sec \theta = \frac{AC}{BC} and \cosec(90^\circ - \theta) = \frac{AC}{BC}. So \sec \theta = \cosec(90^\circ - \theta). The reverse is \cosec \theta = \sec(90^\circ - \theta).
| Ratio of \theta | Equal ratio of (90^\circ - \theta) |
|---|---|
| \sin \theta | \cos(90^\circ - \theta) |
| \cos \theta | \sin(90^\circ - \theta) |
| \tan \theta | \cot(90^\circ - \theta) |
| \cot \theta | \tan(90^\circ - \theta) |
| \sec \theta | \cosec(90^\circ - \theta) |
| \cosec \theta | \sec(90^\circ - \theta) |
The chapter states a compact form for angles that sum to 90°. If A + B = 90°, then \sin A = \cos B, \tan A = \cot B, and \sec A = \cosec B.
A common mistake is applying the identity in the wrong direction. When you see \sin 28^\circ, do not change it to \cos 28^\circ. The rule gives \sin 28^\circ = \cos(90^\circ - 28^\circ) = \cos 62^\circ. Replace the angle with its complement, not with itself.
Worked pattern: evaluate \frac{\cos 24^\circ}{\sin 66^\circ} + \frac{\tan 29^\circ}{\cot 61^\circ}. Since \cos 24^\circ = \sin 66^\circ, the first fraction is 1. Since \tan 29^\circ = \cot 61^\circ, the second fraction is 1. The sum is 2.
Triangle questions follow the same rule. With A + B + C = 180^\circ, dividing by 2 gives \frac{A+B}{2} = 90^\circ - \frac{C}{2}. Applying the identity gives \sin\frac{A+B}{2} = \cos\frac{C}{2}. The chapter also proves \sec\frac{A+B}{2} = \cosec\frac{C}{2}.
What to practise from Complementary Angles Class 9 ICSE
Section 25.3 carries seven solved examples that model every question type. Work through them before attempting the exercise. Each example shows the substitution step where an angle is replaced by its complement.
- Seven solved examples in section 25.3 — evaluating ratio expressions, proving identities, and finding an unknown angle.
- EXERCISE 25 — 8 questions — evaluating expressions, proving identities, rewriting ratios in terms of angles between 0° and 45°, and triangle-based proofs.
You can check the official Class 9 syllabus and examination details on the CISCE website.
Complementary Angles Class 9 ICSE frequently asked questions
What are complementary angles in Class 9 ICSE trigonometry?
Two acute angles are complementary when their sum is 90°. For any angle \theta^\circ, the complement is (90 - \theta)^\circ. In this ICSE chapter, that definition pairs each trigonometric ratio with the ratio of its complement.
How do you prove that sin(90° − θ) equals cos θ in Class 9 ICSE?
In a right triangle, both expressions are the same side ratio. \sin(90^\circ - \theta) is perpendicular over hypotenuse at the complement angle, which is exactly how \cos \theta is defined. The two fractions are identical.
What is the relation between tan and cot for complementary angles in this ICSE chapter?
\tan \theta = \cot(90^\circ - \theta) and \cot \theta = \tan(90^\circ - \theta). Tangent and cotangent swap places when you replace an angle with its complement.
How do you solve triangle questions like proving sin((A+B)/2) = cos(C/2)?
In any triangle, A + B + C = 180^\circ. Dividing by 2 gives \frac{A+B}{2} = 90^\circ - \frac{C}{2}. Applying \sin(90^\circ - \theta) = \cos \theta with \theta = \frac{C}{2} gives the result directly.
Reference: ICSE Class 9 Mathematics textbook material, chapter 25: Complementary Angles — CISCE-aligned.