QPaperGen

We use the standard trigonometric identities:


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




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



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


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





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



Therefore,


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