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