Prove the Identity : \[ \frac{\cos x}{1-\sin x} = \frac{1+\cos x+\sin x}{1+\cos x-\sin x} \]

Solution:

\[ \frac{\cos x}{1-\sin x} \]

\[ = \frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)} \]

\[ = \frac{\cos x(1+\sin x)}{1-\sin^2 x} \]

\[ = \frac{\cos x(1+\sin x)}{\cos^2 x} \]

\[ = \frac{1+\sin x}{\cos x} \]

\[ = \frac{(1+\sin x)(1+\cos x)} {\cos x(1+\cos x)} \]

\[ = \frac{1+\cos x+\sin x+\sin x\cos x} {\cos x+\cos^2 x} \]

Using \[ 1-\sin^2 x=\cos^2 x \]

\[ \cos^2 x=1-\sin^2 x \]

\[ \cos x+\cos^2 x = \cos x+1-\sin^2 x \]

\[ = (1+\cos x+\sin x)(1+\cos x-\sin x) \div (1+\cos x+\sin x) \]

\[ = \frac{1+\cos x+\sin x} {1+\cos x-\sin x} \]

Hence proved.