Question

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

Answer 1

$\begin{array}{c}\frac{{\cos }^2\theta }{\left(1-\tan \theta \right)}+\frac{{\sin }^3\theta }{\left(\sin \theta -\cos \theta \right)}=\left(+\sin \theta .\cos \theta \right)\\ \Rightarrow \frac{{\cos }^2\theta }{\left(\cos \theta -\sin \theta \right)}+\frac{{\sin }^3}{\left(\sin \theta -\cos \theta \right)}\\ \Rightarrow \frac{-{\cos }^3\theta +{\sin }^3\theta }{\cos \theta }=\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }\\ \Rightarrow \frac{\sin \theta -\cos \theta }{\left(\sin \theta -\cos \theta \right)}\left[{\sin }^2\theta +\sin \theta -\cos \theta +{\cos }^2\theta \right]\\ \Rightarrow \left\{\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta +\cos \theta \right)\right\}=\\ \Rightarrow 1+s\sin \theta \cdot \cos \theta Ans.\end{array}$