Prove the following : (sin^2(160°)/sin^2 70°) + (sin(180°-θ)/sinθ) = sec^2 20°

Question

Prove the following :

\(\frac {sin^2(160°)}{sin^270°} +\frac{sin(180°-\theta)}{sin\theta}=sec^2 20°\)

(sin²(160°)/sin²70°) + (sin(180°-θ)/sinθ) = sec²20°

Answer 1

  = 1 + tan²20^∘

  = sec² 20^∘ = R.H.S.