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

Sagot :

Answer:

[tex] \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 }} [/tex]

#hence proved

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