Question

(i) If sinθ + cosθ = p and secθ + cosecθ = q then prove that q(p² – 1) = 2p

(ii) If sinθ(1 + sin² θ) = cos² θ, then prove that cos⁶ θ – 4cos⁴ θ + 8cos² θ = 4

Answer 1

(i) p = sinθ + cosθ  

(ii) Given sinθ(1 + sin² θ) = cos² θ

Substitute sin² θ = 1 – cos² θ and take cos θ = c

Squaring (1) on both sides we get

sin² θ(1 + sin² θ)² = cos⁴ θ

(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 + 8cos² θ = 4 = RHS

∴ Hence proved