Question

Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos²θ

Answer 1

L.H.S.

= `(tan(90° - θ)cotθ)/("cosec"^2 θ)` 

= `(cotθ xx cotθ)/("cosec^2 θ)`

= `(cot^2 θ)/("cosec"^2 θ)`

= `((cos^2 θ)/(sin^2 θ))/((1)/(sin^2 θ)`

= cos²θ
= R.H.S.