**NCERT solutions for class 10 Maths**

**Chapter 8**

## Introduction to Trigonometry

**Exercise 8.4**

**Ex 8.4 Question 1.**

**Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.**

**Solution:**

**Ex 8.4 Question 2.**

**Write all the other trigonometric ratios of ∠A in terms of sec A.
**

**Solution:**

**Ex 8.4 Question 3.**

**Evaluate:**

**(i) (sin ^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)**

(ii) sin 25° cos 65° + cos 25° sin 65°

**Solution:**

**4. Choose the correct option. Justify your choice.
(i) 9 sec ^{2}A – 9 tan^{2}A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) – 1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A**

**(iv) 1+tan ^{2}A/1+cot^{2}A = **

** (A) sec ^{2 }A (B) -1 (C) cot^{2}A (D) tan^{2}A**

**Solution:**

**5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.**

**(i) (cosec θ – cot θ) ^{2 }= (1-cos θ)/(1+cos θ)**

**(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A**

**(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ**

** [Hint : Write the expression in terms of sin θ and cos θ]**

**(iv) (1 + sec A)/sec A = sin ^{2}A/(1-cos A) **

** [Hint : Simplify LHS and RHS separately]**

**(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec ^{2}A = 1+cot^{2}A.**

**(vii) (sin θ – 2sin ^{3}θ)/(2cos^{3}θ-cos θ) = tan θ**

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

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

[Hint : Simplify LHS and RHS separately]

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} =^{ }tan^{2}A

**Solution:**