Question

Prove that `("cosec"  θ)/("cosec"  θ - 1) + ("cosec"  θ)/("cosec"  θ + 1) = 2 sec^2 θ`

Answer 1

LHS = `("cosec"  θ)/("cosec"  θ - 1) + ("cosec"  θ)/("cosec"  θ + 1)`

= `"cosec"  θ [1/("cosec"  θ - 1) + 1/("cosec"  θ + 1)]`

= `"cosec"  θ [("cosec"  θ + 1 + "cosec"  θ - 1)/(("cosec"  θ - 1)("cosec"  θ + 1))]`

= `"cosec"  θ ((2  "cosec"  θ)/("cosec"^2θ - 1))`

= `(2  "cosec"^2θ)/("cosec"^2θ - 1)`

= `(2  "cosec"^2θ)/(cot^2θ)`   ...[1 + cot²θ = cosec²θ]

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

= `2/(cos^2θ)`

= 2 sec²θ

= RHS

Hence Proved.