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Question
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
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Solution
LHS = `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ)`
= `((cos θ + sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ + sin θ) + ((cos θ - sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ - sin θ)`
= 1 - sin θ cos θ + 1 + sin θ cos θ
= 2
= RHS
Hence proved.
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