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

\[\begin{aligned} & \mathrm{LHS}=\frac{\cos ^{2} \theta}{1-\tan \theta}+\frac{\sin ^{3} \theta}{\sin \theta-\cos \theta} \\ & =\frac{\cos ^{2} \theta}{1-\dfrac{\sin \theta}{\cos \theta}}+\frac{\sin ^{3} \theta}{\sin \theta-\cos \theta} \\ & =\frac{\cos ^{3} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{3} \theta}{\cos \theta-\sin \theta} \\ & =\frac{\cos ^{3} \theta-\sin 3 \theta}{\cos \theta-\sin \theta} \\ & =\frac{(\cos \theta-\sin \theta)\left(\cos ^{2} \theta+\sin ^{2} \theta+\cos \theta \sin \theta\right)}{(\cos \theta-\sin \theta)} \\ & =1+\cos \theta \sin \theta=\mathrm{RHS} \end{aligned} \]