Answer 1

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