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Question
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
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Solution
LHS = `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
= `((cos^3A + sin^3A)(cosA - sinA) + (cos^3A - sin^3A)(cosA + sinA))/(cos^2A - sin^2A)`
= `(cos^4A - cos^3AsinA + sin^3AcosA - sin^4A + cos^4A + cos^3AsinA - sin^3AcosA = sin^4A)/(cos^2A - sin^2A)`
= `(2(cos^4A - sin^4A))/(cos^2A - sin^2A) = (2(cos^2A + sin^2A)(cos^2A - sin^2A))/((cos^2A - sin^2A)) = 2(cos^2A + sin^2A)`
= `2(∵ cos^2A + sin^2A = 1)`
OR
= `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
= `((cosA + sinA)(cos^2A + sin^2A - cosAsinA))/((cosA + sinA)) + ((cosA - sinA)(cos^2A + sin^2A + cosAsinA))/((cosA - sinA))` (∵ a3 ± b3 = (a ± b)(a2 + b2 ± ab))
= `(cos^2A + sin^2A - cosAsinA) + (cos^2A + sin^2A + cosAsinA)`
= `1 - cosAsinA + 1 + cosAsinA ` (∵ `cos^2A + sin^2A = 1`)
= 2
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