Question

Prove that [1 + cot θ – sec (θ + π/2)] [1 + cot θ + sec (θ + π/2)] = 2 cot θ

Answer 1

L.H.S.
[1 + cot θ – sec (θ + π/2)] [1 + cot θ + sec (θ + π/2)]

= [1 + cot θ + cosec θ] [1+ cot θ – cosec θ]

= (1 + cot)² – cosec²θ

= 1 + cot²θ + 2 cotθ – cosec²θ

= 1 + 2 cot θ – (cosec2θ – cot2θ)

= 1+2 cot θ – 1

= 2 cot θ = R.H.S.

Hence Proved.