Prove the Identity : \[ \frac{\tan^3 x}{1+\tan^2 x} + \frac{\cot^3 x}{1+\cot^2 x} = \frac{1-2\sin^2 x\cos^2 x}{\sin x\cos x} \]

Solution:

\[ \frac{\tan^3 x}{1+\tan^2 x} + \frac{\cot^3 x}{1+\cot^2 x} \]

Using \[ 1+\tan^2 x=\sec^2 x \quad \text{and} \quad 1+\cot^2 x=\cosec^2 x \]

\[ = \frac{\tan^3 x}{\sec^2 x} + \frac{\cot^3 x}{\cosec^2 x} \]

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

\[ = \frac{\sin^4 x+\cos^4 x}{\sin x\cos x} \]

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

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

Hence proved.