Prove the following trigonometric identities.

We need to prove $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$

Solving the L.H.S, we get

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$

$=\frac{\frac{\cos A}{\cos A-\sin A}}{\frac{\cos A}{\sin A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin }}$

$=\frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}$

$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$

$=\frac{(\cos A+\sin A)(\cos A-\sin A)}{\cos A-\sin A}\left[\right.$ using $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\cos A+\sin A$

= RHS

Hence proved.