Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\cos }^2\theta }{1-\tan \theta }+\frac{{\sin }^3\theta }{\sin \theta -\cos \theta } \\ =\frac{{\cos }^3\theta }{\cos \theta -\sin \theta }+\frac{{\sin }^3\theta }{\sin \theta -\cos } \\ =\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta } \\ =\frac{\left(\cos \theta -\sin \theta \right)\left({\cos }^2\theta +{\sin }^2\theta +\sin \theta \cos \theta \right)}{\left(\cos \theta -\sin \theta \right)} \end{gathered}$$

^{$$\begin{gathered} ={\cos }^2\theta +{\sin }^2\theta +\sin \theta \cos \theta \\ =1+\sin \theta \cos \theta \\ =\mathrm{RHS} \end{gathered}$$}

Hence, proved.