Question

Prove that sin theta / (cot theta + cosec theta) = 2 + [sin theta / (cot theta - cosec theta)]

Answer 1

Step-by-step explanation:

*LHS*= sinθ/ cotθ + cosecθ = sinθ / (cosθ/ sinθ+ 1/sinθ)=

sin²θ/ (cosθ+1) = (1- cos²θ)/1+cosθ = (1-cosθ)(1+cosθ)/1+cosθ

= 1 - cosθ

*RHS* = 2 + sinθ/cotθ - cosecθ = 2 + sinθ/ (cosθ-1)/sinθ

= 2 + sin²θ/(cosθ -1) = 2 + (1+ cosθ)(1-cosθ)/-1×(1-cosθ)

= 2 -(1+ cosθ) = 1 - cosθ

Hence, LHS=RHS