10th Maths Book Back Question and Answers – Chapter 6 Exercise 6.1:

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Samacheer Kalvi 10th Maths Book Back Answers – Ex 6.1 Trigonometry

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Chapter 6

Exercise 6.1 Trigonometry

1. Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan⁴θ + tan²θ = sec⁴θ – sec²θ
Solution:

2. Prove the following identities

Solution:

3. Prove the following identities

Solution:

4. Prove the following identities
(i) sec⁶θ = tan⁶θ + 3tan²θ sec²θ + 1
(ii) (sinθ + secθ)² + (cosθ + cosecθ)²= 1 + (secθ + cosecθ)²
Solu:

(i) L.H.S = sec⁶θ = (sec²θ)³ = (1 + tan²θ )³ = (tan²θ + 1)³
(a + b)³ = a³ + 3a²b + 3ab² + b³
= (tan²θ)³ + 3(tan²θ)² × 1 + 3 × tan²θ × 1² + 1
= tan⁶θ + 3tan²θ × (sec²θ – 1) + 3tan²θ + 1
= tan⁶θ + 3tan²θsec²θ – 3tan²θ + 3tan²θ +1
= tan⁶θ + 3tan²θ sec²θ + 1 = R.H.S

(ii) L.H.S = (sinθ + secθ )² + (cosθ + cosecθ)²
= sin²θ + 2sinθ secθ + sec²θ + cos²θ + 2cosθ cosecθ+ cosec²θ

5. Prove the following identities

Solution:

(ii)

6. Prove the following identities

7.

Solution:

8.

Solution:
(i) LHS:

9. (i) If sinθ + cosθ = p and secθ + cosecθ = q then prove that q(p² – 1) = 2p
(ii) If sinθ(1 + sin²θ) = cos²θ, then prove that cos⁶θ – 4cos⁴θ + 8cos²θ = 4
Solution:

(ii) Given sinθ(1 + sin²θ) = cos²θ
Substitute sin²θ = 1 – cos²θ and take cos θ = c
squaring (1) on both sides we get
sin²θ(1 + sin²θ)² = cos⁴θ
(1 – c²)(1 + 1 – c²) = c⁴
(1 – c²)(2 – c²)² = c⁴
(1 – c²)(4 + c⁴– 4c²) = c⁴
4 + c⁴– 4c² – 4c² – c⁶ + 4c⁴ = c⁴
 -c^{6 +} 4c⁴ – 8c² = -4
c⁶ – 4c⁴ + 8c² = -4
ie cos 6θ – 4cos 4θ + 8cos²θ = 4 = RHS
∴ Hence proved

10.

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