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प्रश्न
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
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उत्तर
We have prove that
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
We know that `sin^2 A + cos^2 A = 1`
So,
(2 + cot A + tan A)(sin A - cos A)
`= (1 + cos A/sin A + sin A/cos A)(sin A - cos A)`
`= ((sin A cos A + cos^2 A + sin^2 A)/(sin A cos A)) (sin A - cos A)`
`= ((sin A cos A + 1)/(sin A cos A))(sin A - cos A)`
`= ((sin A - cos A)(sin A cos A + 1))/(sin A cos A)`
`= (sin^2 A cos A + sin A - cos^2 A sin A - cos A)/(sin A cos A)`
`= ((sin^2 A cos A - cos A) + (sin A - cos^2 A sin A))/(sin A cos A)`
`= (cos A (sin^2 A - 1)+ sin A (1 - sin^2 A))/(sin A cos A)`
`= (cos A (-cos^2 A) + sin A (sin^2 A))/(sin A cos A)`
`= (-cos^3 A + sin^3 A)/(sin A cos A)`
`= (sin^3 A - cos^3 A)/(sin A cos A)`
`= (sin^2 A)/cos A - cos^2 A/sin A`
`= sin A/cos A sin A - cos A/sin A cos A`
`= tan A sin A - cot A cos A`
= sin A tan A - cos A cot A
Now
`sec A/(cosec^2 A) - (cosec A)/sec^2 A = (1/cos A)/(1/sin^2 A) - (1/sin A)/(1/cos^2 A)`
`= sin^2 A/cos A - cos^2 A/sin A`
`= sin A sin A/cos a - cos A cos A/sin A`
= sin A tan A - cos A cot A
Hence proved.
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