# Question #f61d0

May 20, 2017

$L H S = \frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x + \cos x}$

$= \frac{\left(\sin x - \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x + \sin x \cos x\right)}{\sin x + \cos x}$

$= \frac{{\left(\sin x - \cos x\right)}^{2} \left(1 + \sin x \cos x\right)}{\left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}$

$= \frac{\left({\sin}^{2} x + {\cos}^{2} x - 2 \sin x \cos x\right) \left(1 + \sin x \cos x\right)}{\left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}$

$= \frac{\left(1 - 2 \sin x \cos x\right) \left(1 + \sin x \cos x\right)}{{\sin}^{2} x - {\cos}^{2} x}$

$= \frac{\left(\frac{1}{\sin} ^ 2 x - \frac{2 \sin x \cos x}{\sin} ^ 2 x\right) \left(1 + \sin x \cos x\right)}{{\sin}^{2} \frac{x}{\sin} ^ 2 x - {\cos}^{2} \frac{x}{\sin} ^ 2 x}$

$= \frac{\frac{1}{\sin} ^ 2 x - \frac{2 \sin x \cos x}{\sin} ^ 2 x + \frac{\sin x \cos x}{\sin} ^ 2 x - 2 {\cos}^{2} x}{1 - {\cot}^{2} x}$

$= \frac{\frac{1}{\sin} ^ 2 x - 2 \cot x + \cot x - 2 {\cos}^{2} x}{1 - {\cot}^{2} x}$

$= \frac{{\csc}^{2} x - \cot x - 2 {\cos}^{2} x}{1 - {\cot}^{2} x} = R H S$

Proved