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

SOLUTION:

(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
Sustitute sin2θ = 1 – co2θ and take cos θ = c
squaring (1) on bothsides 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
 -c6 + 4c4 – 8c2 = -4
c6 – 4c4 + 8c2 = -4
ie cos 6θ – 4cos 4θ + 8cos2θ = 4 = RHS
∴ Hence proved

10.