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