Question

Prove the following identities
\[\frac{1 - \sin x \cos x}{\cos x \left( \sec x - cosec x \right)} \cdot \frac{\sin^2 x - \cos^2 x}{\sin^3 x + \cos^3 x} = \sin x\]

Answer 1

\[\text{ LHS }= \frac{1 - \sin x \cos x}{\cos x \left( \sec x - cosec x \right)} \times \frac{\sin^2 x - \cos^2 x}{\sin^3 x + \cos^3 x}\]
\[ = \frac{1 - \sin x \cos x}{\cos x \left( \frac{1}{\cos x} - \frac{1}{\sin x} \right)} \times \frac{\left( \sin x \right)^2 - \left( \cos x \right)^2}{\left( \sin x \right)^3 + \left( \cos x \right)^3}\]
\[ = \frac{1 - \sin x \cos x}{\cos x \left( \frac{\sin x - \cos x}{\cos x \sin x} \right)} \times \frac{\left( \sin x + \cos x \right)\left( \sin x - \cos x \right)}{\left( \sin x + \cos x \right)\left[ \left( \sin x \right)^2 + \left( \cos x \right)^2 - \sin x \cos x \right]}\]
\[ = \frac{\sin x\left( 1 - \sin x \cos x \right)}{\left( \sin x - \cos x \right)} \times \frac{\left( \sin x + \cos x \right)\left( \sin x - \cos x \right)}{\left( \sin x + \cos x \right)\left[ \left( \sin x \right)^2 + \left( \cos x \right)^2 - \sin x \cos x \right]}\]
\[ = \frac{\sin x\left( 1 - \sin x \cos x \right)}{1} \times \frac{1 \times 1}{\left[ \sin^2 x + \cos^2 x - \sin x \cos x \right]}\]
\[ = \sin x\left( 1 - \sin x \cos x \right) \times\frac{1}{\left( 1 - \sin x \cos x \right)}\]
\[ = \sin x\]
 = RHS
Hence proved.