Question

Prove : $\frac{{\cos }^2\theta }{1-\tan \theta }+\frac{{\sin }^3\theta }{\sin \theta -\cos \theta }=1+\sin \theta \cos \theta$

Answer 1

LHS

$=\frac{co{s}^2\theta }{1-tan\theta }+\frac{si{n}^3\theta }{sin\theta -cos\theta }$

$=\frac{co{s}^3\theta }{cos\theta -sin\theta }-\frac{si{n}^3\theta }{cos\theta -sin\theta }$

$=\frac{co{s}^3\theta -si{n}^3\theta }{cos\theta -sin\theta }$

$=\frac{(cos\theta -sin\theta )(co{s}^2\theta +si{n}^2\theta +cos\theta sin\theta )}{cos\theta -sin\theta }$

$=1+sin\theta cos\theta$ = RHS