Prove the Identity : \[ \left( \frac{1}{\sec^2 x-\cos^2 x} + \frac{1}{\cosec^2 x-\sin^2 x} \right) \sin^2 x\cos^2 x = \frac{1-\sin^2 x\cos^2 x} {2+\sin^2 x\cos^2 x} \]

Solution:

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

\[ = \left( \frac{\cos^2 x}{1-\cos^4 x} + \frac{\sin^2 x}{1-\sin^4 x} \right) \sin^2 x\cos^2 x \]

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

\[ = \frac{\cos^4 x}{1+\cos^2 x} + \frac{\sin^4 x}{1+\sin^2 x} \]

\[ = \frac{\cos^4 x(1+\sin^2 x)+\sin^4 x(1+\cos^2 x)} {(1+\cos^2 x)(1+\sin^2 x)} \]

\[ = \frac{\sin^4 x+\cos^4 x+\sin^2 x\cos^2 x} {2+\sin^2 x\cos^2 x} \]

Using \[ \sin^4 x+\cos^4 x = (\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x \]

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

Therefore,

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

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

Hence proved.