Prove the following: i. tan θ + tan θ = sec θ – sec θ ii. 1/(1−sinθ) + 1/(1 + sinθ) = 2 sec^2θ

Question

Prove the following:

i. tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

ii. 1/(1−sinθ) + 1/(1 + sinθ) = 2 sec²θ

Answer 1

i. L.H.S. = tan⁴ θ + tan² θ 

= tan² θ (tan² θ + 1) 

= tan² θ. sec² θ …[∵ 1 + tan² θ = sec² θ] 

= (sec² θ – 1) sec² θ …[∵ tan² θ = sec² θ – 1] 

= sec⁴ θ – sec⁴ θ 

= R.H.S. 

∴ tan⁴ θ + tan² θ = sec⁴ θ – sec² θ