Prove the following trigonometric identity :

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\cos^2\theta}{1 - \tan\theta} + \frac{\sin^2\theta}{1 - \cot\theta} \\[1em] \Rightarrow \frac{\cos^2\theta}{1 - \frac{\sin\theta}{\cos\theta}} + \frac{\sin^2\theta}{1 - \frac{\cos\theta}{\sin\theta}} \\[1em] \Rightarrow \frac{\cos^2\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} + \frac{\sin^2\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}} \\[1em] \Rightarrow \frac{\cos^3\theta}{\cos\theta - \sin\theta} + \frac{\sin^3\theta}{\sin\theta - \cos\theta} \\[1em] \Rightarrow \frac{\cos^3\theta}{\cos\theta - \sin\theta} - \frac{\sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \frac{\cos^3\theta - \sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \frac{(\cos\theta - \sin\theta)(\cos^2\theta + \cos\theta,\sin\theta + \sin^2\theta)}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \cos^2\theta + \sin^2\theta + \cos\theta,\sin\theta \\[1em] \Rightarrow 1 + \cos\theta,\sin\theta \qquad [\because \sin^2\theta + \cos^2\theta = 1]\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\mathbf{\frac{\cos^2\theta}{1 - \tan\theta} + \frac{\sin^2\theta}{1 - \cot\theta} = 1 + \cos\theta,\sin\theta}$.