Concise Selina Solutions for Class 9 Maths Chapter 25 - Complementary Angles

Selina Solutions are useful for students as it helps them in scoring high marks in the examination. The Selina Solutions given in this page contains detailed step-by-step explanations to all the problems that come under Chapter 25, Complementary Angles, of the Class 9 Selina Textbook. This chapter is a continuation of Chapter 23 and 24 of class 9 Selina book. This chapter deals with the concept of Complementary Angles.

These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. By understanding the concepts used in Selina Solutions for Class 9 Maths, students will be able to clear all their doubts related to “Complementary Angles” of triangles.

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Exercise 25

1. Evaluate:

(i) cos 22^o/sin 68^o

(ii) tan 47^o/cot 43^o

(iii) sec 75^o/cosec 15^o

(v) sin² 40^o – cos² 50^o

(vi) sec² 18^o – cosec² 72^o

(vii) sin 15^o cos 15^o – cos 75^o sin 75^o

(viii) sin 42^o cos 48^o – cos 42^o sin 48^o

Solution:

(i) According to the question we have,

cos 22^o/sin 68^o = cos (90^o – 68^o)/ sin 68^o

by using complementary angle identity,

= sin 68^o / sin 68^o

= 1

(ii) According to the question we have,

tan 47^o/cot 43^o = tan (90^o – 43^o)/ cot 43^o

by using complementary angle identity,

= cot 43^o / cot 43^o

= 1

(iii) According to the question we have,

sec 75^o/cosec 15^o = sec (90^o – 15^o)/ cosec 15^o

by using complementary angle identity,

= cosec 15^o / cosec 15^o

= 1

(iv) According to the question we have,

= 1 + 1

= 2

(v) sin² 40^o – cos² 50^o

= sin² (90^o – 50^o) – cos² 50^o

= cos² 50^o – cos² 50^o

= 0

(vi) sec² 18^o – cosec² 72^o

= [sec (90^o – 72^o)]² – cosec² 72^o

= cosec² 72^o – cosec² 72^o

= 0

(vii) sin 15^o cos 15^o – cos 75^o sin 75^o

= sin (90^o – 75^o) cos 15^o – cos 75^o sin (90^o – 15^o)

= cos 75^o cos 15^o – cos 75^o cos 15^o

= 0

(viii) sin 42^o sin 48^o – cos 42^o cos 48^o

= sin (90^o – 48^o) sin 48^o – cos (90^o – 48^o) cos 48^o

= cos 48^o sin 48^o – sin 48^o cos 48^o

= 0

2. Evaluate:

(i) Sin (90^o – A) sin A – cos (90^o – A) cos A

(ii) sin² 35^o – cos² 55^o

(iii)

(iv)

(v) cos² 25^o – sin²65^o – tan²45^o

(vi)

Solution:

(i) Given Sin (90^o – A) sin A – cos (90^o – A) cos A

= cos A sin A – sin A cos A

= 0

(ii) Given sin² 35^o – cos² 55^o

= sin² 35^o – [cos (90^o – 35^o)]²

= sin² 35^o – sin² 35^o

= 0

(iii) According to the question we have,

On simplifying we get

= 1 + 1 – 2

= 2 – 2

= 0

(iv) According to the question we have

On simplifying we get

= 2 – 1

= 1

(v) According to the question we have

Cos² 25^o – sin² 65^o – tan² 45^o

= [cos (90^o – 65^o)]² – sin² 65^o – (tan² 45^o)²

= sin² 65^o – sin² 65^o – 1²

= 0 – 1

= -1

(vi) According to the question we have,

On simplifying we get

= 1² + 1² – 2 (½)

= 1 + 1 – 1

= 1

3. Show that:

(i) tan 10° tan 15° tan 75° tan 80° = 1

(ii) sin 42° sec 48° + cos 42° cosec 48° = 2

Solution:

(i) L.H.S.

= tan 10° tan 15° tan 75° tan 80°

= tan (90° – 80°) tan (90° – 75°) tan 75° tan 80°

= cot 80° cot 75 ° tan 75° tan 80°

= (cot 80° tan 80°) (cot 75° tan 75°)

= (1)(1)

= 1

= R.H.S.

(ii) consider LHS

On simplifying we get,

= 1 + 1

= 2

= RHS

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

(i) sin 59°+ tan 63°

(ii) cosec 68°+ cot 72°

(iii) cos 74°+ sec 67°

Solution:

(i) Given sin 59°+ tan 63°

It can be written as

sin 59°+ tan 63° = sin (90 – 31)^o + tan(90 – 27)^o

= cos 31^o + cos 27^o

(ii) Given cosec 68°+ cot 72°

It can be written as

cosec 68°+ cot 72° = = cosec (90 – 22)^o + cot (90 – 18)^o

= sec 22^o + tan 18^o

(iii) Given cos 74°+ sec 67°

It can be written as

cos 74°+ sec 67° = cos (90 – 16)^o + sec(90 – 23)^o

= sin 16^o + cosec 23^o

5. Question 5

For triangle ABC, show that:

(i) 

(ii) 

Solution:

(i) We know that for △ ABC

∠ A + ∠ B + ∠ C = 180^o

∠ B + ∠ A = 180^o = ∠ C

(∠ B + ∠ A) /2 = 90^o – ∠ C/2

Sin [(A+B)/2] = sin (90^o – C/2)

= cos (C/2)

(ii) We know that for △ ABC

∠ A + ∠ B + ∠ C = 180^o

∠ B + ∠ C = 180^o = ∠ A

(∠ B + ∠ C) /2 = 90^o – ∠ A/2

tan [(B + C)/2] = tan (90^o – A/2)

= cot (A/2)

6. Evaluate:

(i) 

(ii) 3 cos 80^o cosec 10^o + 2 sin 59^o sec 31^o

(iii) 

(iv) tan (55^o – A) – cot (35^o + A)

(v) cosec (65^o + A) – sec (25^o – A)
(vi) 

(vii) 

(viii) 

(ix) 14 sin 30^o + 6 cos 60^o – 5 tan 45^o

Solution:

(i) Given

(ii) Given 3 cos 80^o cosec 10^o + 2 sin 59^o sec 31^o

= 3 cos (90^o – 19^o) cosec 10^o + 2 sin (90^o – 31^o) sec 31^o

= 3 sin 10^o cosec 10^o + 2 cos 31^o sec 31^o

= 3 (1) + 2 (1)

= 3 + 2

= 5

(iii) Given

On simplifying we get

= 1 + 1

= 2

(iv) Given tan (55^o – A) – cot (35^o + A)

It can be written as

tan (55^o – A) – cot (35^o + A) = tan [90 – (35^o + A) – cot (35^o + A)

= cot (35^o + A) – cot (35^o + A)

= 0

(v) Given cosec (65^o + A) – sec (25^o – A)

It can be written as

cosec (65^o + A) – sec (25^o – A) = cosec [90^o – (25^o – A) – sec (25^o – A)

= sec (35^o + A) – sec (25^o – A)

= 0

On simplifying we get,

= 2 – 1 – 1

= 0

(vii) Given

On simplifying we get

= 1 – 2

= -1

(viii) Given

On simplifying we get

= 1 + 1 – 2

= 2 – 2

= 0

(ix) Given 14 sin 30^o + 6 cos 60^o – 5 tan 45^o

It can be written as

14 sin 30^o + 6 cos 60^o – 5 tan 45^o = 14 (½) + 6 (½) – 5 (1)

= 7 + 3 – 5

= 10 – 5

= 5

7. A triangle ABC is right angled at B. Find the value of

Solution:

Since triangle ABC is a right-angled triangle, and right angle at B

Therefore, A + C = 90^o

On simplifying we get

= 1 – 1

= 0

8. 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° – 3A).sec 77° = 1

Solution:

(i) Given sin (90° – 3A). cosec 42° = 1

On rearranging we get

sin (90° – 3A) = 1/cosec 42^o

cos 3A = 1/cosec (90^o – 48^o)

cos 3A = 1/sec 48^o

cos 3A = cos 48^o

3A = 48^o

Therefore, A = 16^o

(ii) Given cos (90° – 3A).sec 77° = 1

cos (90° – 3A) = 1/ sec 77° 

sin 3A = 1/ sec (90^o – 12^o)

sin 3A = 1/cosec 12^o

sin 3A = sin 12^o

3A = 12^o

A = 3^o

Selina Solutions for Class 9 Maths Chapter 25- Complementary Angles

Chapter 25, Complementary Angles, is composed of 1 exercise and the solutions given here contain answers to all the questions present in this exercise. Let us have a look at some of the topics that are being discussed in this chapter.

25.1 Introduction

25.2 Concept of trigonometric ratios of complementary angles

25.3 Complementary angles for sine and cosine

Selina Solutions for Class 9 Maths Chapter 25- Complementary Angles

In Chapters 23 and 24, students were taught about different trigonometric ratios and the evaluation of different trigonometric expressions. On the other hand, chapter 25 discusses and uses the concept of trigonometric ratios of complementary angles along with their applications. Read and learn the 25th chapter of Selina textbook to get up to date with the concepts related to Complementary Angles that are taught in the class. Learn the Selina Solutions for Class 9 effectively to score high in the examination.