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प्रश्न
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Prove the following identities:
`(sin A)/((sec A + tan A - 1)) + (cos A)/(("cosec" A + cot A - 1)) = 1`
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उत्तर
LHS = `(sec A)/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1)`
= `(sin A)/(1/cos A + sin A/cos A - 1) + cos A/(1/sin A + cos A/sin A - 1)`
= `(sin A/(1 + sin A - cos A))/cos A + ((cos A)/(1 + cos A - sin A))/(sin A)`
= `(sin A.cos A)/(1 + sin A - cos A) + (sin A. cos A)/(1 + cos A - sin A)`
= `(sin A. cos A( 1 + cos A - sin A + 1 + sin A - cos A))/([ 1 + (sin A - cos A)][1 - (sin A - cos A)])`
= `(2sin A. cos A)/((1)^2 - (sin A - cos A)^2)`
= `(2sin A. cos A)/(1 - (sin^2 A + cos^2 A - 2 sin A.cos A))`
= `(2 sin A. cos A)/(1 - 1 + 2 sin A. cos A)`
= `2/2 = 1`
= RHS
Hence proved.
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