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Question
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
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Solution
L.H.S. = `(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA)`
= `((sinA + cosA)^2 + (sinA - cosA)^2)/((sinA - cosA)(sinA + cosA))`
= `(sin^2A + cos^2A + 2sinAcosA + sin^2A + cos^2A - 2sinA cosA)/(sin^2A - cos^2A)`
= `(2(sin^2A + cos^2A))/(sin^2A - cos^2A)`
= `2/(sin^2A - cos^2A)` ...[sin2A + cos2A = 1]
= `2/(sin^2A - cos^2A)`
= `2/(sin^2A - (1 - sin^2A))`
= `2/(2sin^2A - 1)` = R.H.S.
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