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प्रश्न
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
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उत्तर
L.H.S = `cot^2"A"[(sec"A" - 1)/(1 + sin "A")] + sec^2"A"[(sin"A" - 1)/(1 + sec"A")]`
= `(cot^2"A"(sec"A" - 1)(sec"A" + 1) + sec^2"A"(sin"A" - 1)(sin"A" + 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A"(sec^2"A" - 1) + sec^2"A"(sin^2"A" - 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx tan^2"A" + sec^2"A"( - cos^2"A"))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx 1/cot^2"A" - sec^2"A" xx 1/sec^2"A")/((1 + sin"A")(1 + sec"A"))`
= `(1 - 1)/((1 + sin"A")(1+ sec"A"))`
= `0/((1 + sin"A")(1 + sec"A"))`
= 0
L.H.S = R.H.S
Hence it is proved.
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