Prove the
following identity:
sin cube A + cos cube A divided by sin A + cos A
= 1 - sin A cos A
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Answer:
(sin^3A+Cos^3A) /(sinA+ cosA) =(sinA+cosA) (sin^2A+cos^2A-sinAcosA) /(sinA+cosA) =1-sinAcosA
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