QPaperGen

We use the following identities:


\[
\sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]




\[
\cos^2 A - \sin^2 A = \cos 2A \quad \Rightarrow \quad \sin^2 A - \cos^2 A = -\cos 2A
\]



Now, consider the left-hand side (LHS):


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





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



Therefore,


\[
\text{LHS} = \text{RHS}
\quad \boxed{\text{Hence proved.}}
\]