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Question
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Solution
LHS = `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A)`
= `((sin A + cos A)^2 + (sin A - cos A)^2)/((sin A - cos A)(sin A + cos A))`
= `(sin^2 A + cos^2 A + 2 sin Acos A + sin^2 A + cos^2 A - 2sin A. cos A)/(sin^2 A - cos^2 A)`
= `(2(sin^2A + cos^2 A))/(sin^2 A - cos^2 A)`
= `(2 xx 1)/(sin^2 A - (1- sin^2 A)`
= `2/(sin^2 A - 1+ sin^2 A)`
= `2/(2 sin^2 A - 1)`
= RHS
Hence proved.
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Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
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sec θ = `square`
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∴ L.H.S. = R.H.S.
