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