ICSE Class 9 Maths Chapter 25 Complementary Angles

25 - Complementary Angles

Introduction

Till now, we have studied trigonometric ratios of standard angles such as: 0°, 30°, 45°, 60° and 90°. Evaluation of trigonometric expressions, involving the trigonometric ratios of standard angles is also done.

In this chapter, we shall be discussing and using the concept of trigonometric ratios of complementary angles and with their applications.

Concept Of Trigonometric Ratios Of Complementary Angles

1. Two acute angles are said to be complementary, if their sum is 90°.

(i) 30° and 60° are complementary as 30° + 60° = 90°.

(ii) 70° and 20° are complementary as 70° + 20° = 90°.

(iii) x° and (90 - x)° are complementary as x° + (90 - x)° = 90°.

2. (i) For angle 48°, its complement = 90° - 48° = 42°.

(ii) For angle θ°, its complement = (90 - θ)°.

Teacher's Note

Complementary angles are like two pieces of a puzzle that fit together to make a right angle, similar to how two parts of a task combine to complete the whole job.

Complementary Angles For Sine (sin) And Cosine (cos)

Consider a triangle ABC, in which ∠B = 90°. If ∠ACB = θ; then by the knowledge of geometry, we know ∠CAB = 90° - θ.

\(\sin \theta = \sin \angle ACB = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AB}{AC}\)

\(\cos(90° - \theta) = \cos \angle CAB = \frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC}\)

From I and II, we get: \(\sin \theta = \cos(90° - \theta)\) i.e. \(\cos(90° - \theta) = \sin \theta\).

\(\cos \theta = \cos \angle ACB = \frac{\text{base}}{\text{hypotenuse}} = \frac{BC}{AC}\)

\(\sin(90° - \theta) = \sin \angle CAB = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC}\)

From III and IV, we get: \(\sin(90° - \theta) = \cos \theta\)

Thus,

1. \(\cos(90° - \theta) = \sin \theta\)

2. \(\sin(90° - \theta) = \cos \theta\)

Teacher's Note

The sine of an angle equals the cosine of its complement, just as a student's strength in math complements their understanding of science.

Complementary Angles For Tangent (tan) And Cotangent (cot)

Consider a triangle ABC, in which ∠B = 90°. If ∠ACB = θ; then ∠CAB = 90° - θ.

\(\tan \theta = \tan \angle ACB = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC}\)

\(\cot(90° - \theta) = \cot \angle CAB = \frac{\text{base}}{\text{perpendicular}} = \frac{AB}{BC}\)

\(\tan \theta = \cot(90° - \theta)\) i.e. \(\cot(90° - \theta) = \tan \theta\)

Also,

\(\cot \theta = \cot \angle ACB = \frac{\text{base}}{\text{hypotenuse}} = \frac{BC}{AB}\)

and,

\(\tan(90° - \theta) = \tan \angle CAB = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AB}\)

\(\cot \theta = \tan(90° - \theta)\) i.e. \(\tan(90° - \theta) = \cot \theta\)

Thus: 1. \(\cot(90° - \theta) = \tan \theta\)

(i) \(\cot(90° - 20°) = \tan 20°\)

(ii) \(\cot(90° - 57°) = \tan 57°\)

2. \(\tan(90° - \theta) = \cot \theta\)

(i) \(\tan(90° - 70°) = \cot 70°\)

(ii) \(\tan(90° - 25°) = \cot 25°\)

Teacher's Note

Tangent and cotangent are inverse relationships in complementary angles, similar to how asking a question and finding the answer are complementary actions in learning.

Complementary Angles For Secant (sec) And Cosecant (cosec)

Using the given figure, we find:

\(\sec \theta = \sec \angle ACB = \frac{\text{hypotenuse}}{\text{base}} = \frac{AC}{BC}\)

and,

\(\cosec(90° - \theta) = \cosec \angle CAB = \frac{AC}{BC}\)

\(\sec \theta = \cosec(90° - \theta)\)

i.e. \(\cosec(90° - \theta) = \sec \theta\)

Also,

\(\cosec \theta = \cosec \angle ACB = \frac{AC}{AB}\)

and,

\(\sec(90° - \theta) = \sec \angle CAB = \frac{AC}{AB}\)

\(\cosec \theta = \sec(90° - \theta)\)

i.e. \(\sec(90° - \theta) = \cosec \theta\)

Thus: 1. \(\cosec(90° - \theta) = \sec \theta\)

(i) \(\cosec(90° - 40°) = \sec 40°\)

(ii) \(\cosec(90° - 67°) = \sec 67°\)

2. \(\sec(90° - \theta) = \cosec \theta\)

(i) \(\sec(90° - 35°) = \cosec 35°\)

(ii) \(\sec(90° - 82°) = \cosec 82°\)

1. \(\sin(90° - \theta) = \cos \theta\) and \(\cos(90° - \theta) = \sin \theta\)

2. \(\tan(90° - \theta) = \cot \theta\) and \(\cot(90° - \theta) = \tan \theta\)

3. \(\cosec(90° - \theta) = \sec \theta\) and \(\sec(90° - \theta) = \cosec \theta\)

Teacher's Note

Complementary trigonometric ratios work together like a lock and key, where understanding one automatically helps you understand its complement.

Worked Examples

Example 1

Evaluate:

\(\left(\frac{\cos 47°}{\sin 43°}\right)^2 + \left(\frac{\sin 72°}{\cos 18°}\right)^2 - 2\cos^2 45°\)

Solution

\(\cos 47° = \cos(90° - 43°) = \sin 43°\)

and, \(\sin 72° = \sin(90° - 18°) = \cos 18°\)

\(\left(\frac{\cos 47°}{\sin 43°}\right)^2 + \left(\frac{\sin 72°}{\cos 18°}\right)^2 - 2\cos^2 45°\)

\(= \left(\frac{\sin 43°}{\sin 43°}\right)^2 + \left(\frac{\cos 18°}{\cos 18°}\right)^2 - 2 \times \left(\frac{1}{\sqrt{2}}\right)^2\)

\(= (1)^2 + (1)^2 - 2 \times \frac{1}{2} = 2 - 1 = 1\)

Ans.

If A and B are complimentary angles i.e. if A + B = 90

(i) \(\sin A = \cos B\) and \(\cos A = \sin B\)

(ii) \(\tan A = \cot B\) and \(\cot A = \tan B\)

(iii) \(\sec A = \cosec B\) and \(\cosec A = \sec B\)

Example 2

Evaluate: (i) \(\cosec 82° - \sec 8°\) (ii) \(\sec 70° \sin 20° + \cos 20° \cosec 70°\)

Solution

(i) \(\cosec 82° - \sec 8° = \cosec(90° - 8°) - \sec 8°\)

\(= \sec 8° - \sec 8° = 0\)

Ans.

(ii) \(\sec 70° \sin 20° + \cos 20° \cosec 70°\)

\(= \sec(90° - 20°) \sin 20° + \cos 20° \cosec(90° - 20°)\)

\(= \cosec 20° \sin 20° + \cos 20° \sec 20°\)

\(= \frac{1}{\sin 20°} \times \sin 20° + \cos 20° \times \frac{1}{\cos 20°}\)

\(= 1 + 1 = 2\)

Ans.

Teacher's Note

Working with complementary angles in trigonometry is like using both sides of a coin - each side gives you different information, but together they're complete.

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