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Question
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Solution
L.H.S. = `(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)2(cos^2A - sin^2A))/(cos^2A - sin^2A)`
= 2(cos2 A + sin2 A)
= 2 = R.H.S. ...(∵ cos2 A + sin2 A = 1)
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