Prove the Identity : \[ \frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x} = \sec x\cosec x+1 \]

Solution:

\[ \frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x} \]

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

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

\[ = \frac{\sin^3 x-\cos^3 x} {\sin x\cos x(\sin x-\cos x)} \]

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

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

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

\[ = \sec x\cosec x+1 \]

Hence proved.