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Question
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Solution
LHS = `(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ)`
= `((cos^3θ + sin^3θ)(cosθ - sinθ) + (cos^3θ - sin^3θ)(cosθ + sinθ))/((cosθ + sinθ)(cosθ - sinθ))`
= `(cos^4θ - cos^3θsinθ + sin^3θcosθ - sin^4θ + cos^4θ + cos^3θsinθ - sin^3θcosθ - sin^4θ)/(cos^2θ - sin^2θ)`
= `(2cos^4θ - 2sin^4θ)/(cos^2θ - sin^2θ) = (2(cos^4θ - sin^4θ))/(cos^2θ - sin^2θ)`
= `(2(cos^2θ + sin^2θ)(cos^2θ - sin^2θ))/((cos^2θ - sin^2θ))` = 2(`cos^2θ + sin^2θ`)
= 2 `(∵(cos^2θ + sin^2θ) = 1)`
OR
LHS = `(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ)`
= `((cosθ + sinθ)(cos^2θ + sin^2θ - cosθ sinθ))/(cosθ + sinθ) + ((cosθ - sinθ)(cos^2θ + sin^2θ + cosθsinθ))/((cosθ - sinθ))` (∵ `a^3 ± b^3 = (a ± b)(a^2 + b^2 ± ab`))
= `(cos^2θ + sin^2θ - cosθsinθ) + (cos^2θ + sin^2θ + cosθsinθ)`
= `1 - cosθsinθ + 1 + cosθsinθ` (∵ `cos^2θ + sin^2θ = 1`)
= 2
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