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Question
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
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Solution
We have to prove (sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
We know that `sin^2 A + cos^2 A = 1`
So,
`(sec A − cosec A) (1 + tan A + cot A) = (1/cos A - 1/sin A)(1 + sinA/cos A + cos A/sin A)`
`= ((sin A - cos A)/(sin A cos A))((sin A cos A + sin^2 A + cos^2 A)/(sin A cos 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 + 1))/(sin^2 A cos^2 A)`
`= (sin^2 A cos A + sin A - cos^2 A sin A - cos A)/(sin^2 A cos^2 A)`
`= ((sin^2 A cos A - cos A) + (sin A - cos^2 A sin A))/(sin^2 A cos^2 A)`
`= (cos A(sin^2 A - 1) + sin A (1 - cos^2 A))/(sin^2 A cos^2 A)`
`= (cos A(-cos^2 A) + sin A (sin^2 A))/(sin^2 A cos^2 A)`
`= (-cos^3 A + sin^3 A)/(sin^2 A cos^2 A)`
`= (sin^3 A - cos^3 A)/(sin^2 A cos^2 A)`
`= sin^3 A/(sin^2 A cos^2 A) - cos^3 A/(sin^2 A cos^2 A)`
`= sin A/cos^2 A = cos A/sin^2 A`
`= sin A/cos A 1/cos A - cos A/sin A 1/sin A`
= tan A sec A - cot A cosec A
Hence proved.
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