Prove the Identity : \[ 1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x} = \sin x\cos x \]

Solution:

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

\[ = 1-\frac{\sin^3 x}{\sin x+\cos x} -\frac{\cos^3 x}{\sin x+\cos x} \]

\[ = 1-\frac{\sin^3 x+\cos^3 x}{\sin x+\cos x} \]

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

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

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

\[ =\sin x\cos x \]

Hence proved.