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

Solution:

\[ \frac{1-\sin x\cos x}{\cos x\left(\frac{1}{\cos x}-\frac{1}{\sin x}\right)} \cdot \frac{\sin^2 x-\cos^2 x}{\sin^3 x+\cos^3 x} \]

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

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

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

Using \[ \sin^2 x+\cos^2 x=1 \]

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

\[ =\sin x \]

Hence proved.