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प्रश्न
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
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उत्तर
LHS = `sec^2A + cosec^2A`
= `1/cos^2A + 1/sin^2A = (sin^2A + cos^2A)/(cos^2A.sin^2A)`
= `1/(cos^2A.sin^2A) = sec^2Acosec^2A` = RHS
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We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
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