Question

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

Answer 1

Take LHS of the given equation.

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

$=\frac{{\cos }^3\theta }{\cos \theta -\sin \theta }+\frac{{\sin }^3\theta }{\sin \theta -\cos \theta }$

$=\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }$

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

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

$=RHS$