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

Solution:

\[ \frac{\left(1+\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right) (\sin x-\cos x)} {\frac{1}{\cos^3 x}-\frac{1}{\sin^3 x}} \]

\[ = \frac{ \left( \frac{\sin x\cos x+\cos^2 x+\sin^2 x} {\sin x\cos x} \right) (\sin x-\cos x) } { \frac{\sin^3 x-\cos^3 x} {\sin^3 x\cos^3 x} } \]

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

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

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

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

Hence proved.