Question

Evaluate: $\frac{{\sin }^{3}x}{1+\cos x}+\frac{{\cos }^{3}x}{1-\sin x}$

Answer 1

$si{n}^{3}x=sinxsi{n}^{2}x=sinx(1-co{s}^{2}x)$

so, $\frac{si{n}^{3}x}{1+cosx}=sinx(1-cosx)$

$co{s}^{3}x=cosxco{s}^{2}x=cosx(1-si{n}^{2}x)$

so, $\frac{co{s}^{3}x}{1+sinx}=cosx(1-sinx)$

$\therefore \frac{si{n}^{3}x}{1+cosx}+\frac{co{s}^{3}x}{1+sinx}$

$=sinx(1-cosx)+cosx(1-sinx)$

$=sinx+cosx-2sinxcosx$

$=sinx+cosx-sin2x$