QPaperGen

We use the 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, the left-hand side (LHS) becomes:


\[
\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}
= \frac{2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right)}{2 \cos\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right)}
\]





\[
= \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}
= \frac{\sin(4x)}{\cos(4x)} = \tan(4x)
\]



Hence, LHS = RHS. (Proved)