Question

pls help me to solve this​

Answer 1

Answer:

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Answer 2

$HEYA \: \\ \\ GIVEN \: \: QUESTION \: \: Is \: \\ \\ \frac{ \cos {}^{3} (x) - \sin {}^{3} (x) }{ \cos(x) - \sin(x) } = \frac{1}{2} \frac{(2 + \sin(2x)) }{} \\ \\ \\ lhs \\ \\ \frac{ \cos {}^{3} (x) - \sin {}^{3} (x) }{ \cos(x) - \sin(x) } \\ \\ \frac{( \cos(x) - \sin(x) ) {}^{3} + 3 \sin(x) \cos(x)( \cos(x) - \sin(x) ) }{ \cos(x) - \sin(x) } \\ \\ \\ \frac{( \cos(x) - \sin(x) ) }{( \cos(x) - \sin(x) )} \times \frac{( \cos(x) - \sin(x)) {}^{2} + 3 \sin(x) \cos(x) }{1} \\ \\ \\ \cos {}^{2} (x) + \sin {}^{2} (x) - 2 \sin(x) \cos(x) + 3 \sin(x) \cos(x) \\ \\ \\ 1 + \sin(x) \cos(x) \\ \\ \\ 1 + \frac{1}{2} \frac{2 \sin(x) \cos(x) }{} \\ \\ 1 + \frac{1}{2} \sin(2x) \\ \\ \frac{2 + \sin(2x) }{2} \\ \\ \\ = \frac{1}{2} (2 + \sin(2x) ) \: \: \: hence \: \: proved \: \: \\ \\ \\ NOTE \: \: \: \: \\ \\ 1) \: \: \: a {}^{3} - b {}^{3} = (a - b) {}^{3} + 3ab(a - b) \\ \\ 2) \: \: \: \sin(2x) = 2 \sin(x) \cos(x)$