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Question
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
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Solution
We have to prove `(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
So,
`(cos A cosec A - sin A sec A)/(cos A + sin A) = (cos A 1/sin A - sin A 1/cos A)/(cos A + sin A)`
`= ((cos^2 A - sin^2 A)/(sin A cos A))/(cos A + sin A)`
`= (cos^2 A - sin^2 A)/(sin A cos A(cos A + sin A))`
`= ((cos A - sin A)(cos A + sin A))/(sin A cos A(cos A + sin A))`
`= (cos A - sin A)/(sin A cos A)`
`= cos A/(sin A cos A) - sin A/(sin A cos A)``
`= 1/sin A - 1/cos A``
`= cosec A - sec A`
Hence proved.
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