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Question
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Solution
We have to prove `(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
We know that `sin^2 A = cos^2 A = 1`
`So,
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 = sec A))`
`= (cos^2 A/sin^2 A(1/cos A - 1))/(1 + sin A)`
`= (cos^2 A/sin^2 A (1 - cos A)/(cos A))/(1 + sin A)`
`= (cos A(1 - cos A))/(sin^2 A(1 + sin A))`
`= (cos A (1 - cos A))/((1 - cos^2 A)(1 + sin A))`
`= (cos A (1 - cos A))/((1 - cos A)(1 + cos A)(1 + sin A))`
`= cos A/((1 + cos A)(1 + sin A))`
`= (1/sec A)/((1 + 1/sec A)(1 + sin A))`
`= (1/sec A)/(((sec A + 1)/sec A)) (1 + sin A)`
`= 1/((sec A +1)(1 + sin A))`
Multiplying both the numerator and denominator by (1 - sin A), we have
`= (1 - sin A)/((sec A + 1)(1 + sin A)(1 - sin A))`
`= (1 - sin A)/((sec A + 1)(1 - sin^2 A))`
`= (1 - sin A)/((sec A + 1)cos^2 A)`
`= sec^2 A ((1 - sin A))/((sec A + 1))`
`= sec^2 A ((1 - sin A)/(1 + sec A))`
Hence proved.
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