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Question
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ
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Solution
L.H.S = `1/("cosec" theta - cot theta)`
= `1/("cosec" theta - cot theta) xx ("cosec"theta + cottheta)/("cosec"theta + cottheta)` ......[On rationalising the denominator]
= `("cosec"theta + cottheta)/("cosec"^2theta - cot^2theta)` ......[∵ (a – b)(a + b) = a2 – b2]
= `("cosec"theta cottheta)/1` ......`[(∵ 1 + cot^2θ = "cosec"^2θ),(∴ "cosec"^2θ - cot^2θ = 1)]`
= cosecθ + cotθ
= R.H.S
∴ `1/("cosec" theta - cot theta)` = cosec θ + cot θ
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