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Prove. sin³A-cos³A/sinA-cosA = 1+sinAcosA​

User Tschuege
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Answer:


\frac{ \sin {}^(3) A - \cos {}^(3) A }{ \sin A - \cos A} \\ \\ { \sf{ = \frac{ {( \sin A - \cos A)}^(3) + 3 \sin A \cos A( \sin A - \cos A)}{ \sin A - \cos A} }} \\ \\ = { {( \sin A - \cos A)}^(2) + 3 \sin A \cos A} \\ { \sf{ = ( \sin {}^(2) A + \cos {}^(2) A) - 2 \sin A \cos A + 3\sin A \cos A}} \\ = { \sf{1 +\sin A \cos A }}

#hence proved

User Rhodee
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