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Question
Prove that `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1) = (1 + cos A)/(sin A)`.
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Solution
L.H.S. = `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1)`
= `(cot A + "cosec" A - ("cosec"^2A - cot^2A))/(cot A - "cosec" A + 1)` ...`[(∵ 1 + cot^2A = "cosec"^2A),(∴ "cosec"^2A - cot^2A = 1)]`
= `(cot A + "cosec" A - ("cosec" A + cot A)("cosec" A - cot A))/(cot A - "cosec" A + 1)` ...[∵ a2 – b2 = (a + b) (a – b)]
= `((cot A + "cosec" A)(1 - "cosec" A + cot A))/(cot A - "cosec" A + 1)`
= cot A + cosec A
= `(cos A)/(sin A) + 1/(sin A)`
= `(cos A + 1)/(sin A)`
= R.H.S.
∴ `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1) = (1 + cos A)/(sin A)`
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