QPaperGen

Step 1: Use sum-to-product identities

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

Apply to numerator:
\begin{align*}
\sin 7x + \sin 5x &= 2\sin\left(\frac{7x + 5x}{2}\right)\cos\left(\frac{7x - 5x}{2}\right) = 2\sin(6x)\cos(x) \\
\sin 9x + \sin 3x &= 2\sin\left(\frac{9x + 3x}{2}\right)\cos\left(\frac{9x - 3x}{2}\right) = 2\sin(6x)\cos(3x)
\end{align*}

Apply to denominator:
\begin{align*}
\cos 7x + \cos 5x &= 2\cos(6x)\cos(x) \\
\cos 9x + \cos 3x &= 2\cos(6x)\cos(3x)
\end{align*}

Step 2: Substitute into the expression
\[
\text{LHS} = \frac{2\sin(6x)\cos(x) + 2\sin(6x)\cos(3x)}{2\cos(6x)\cos(x) + 2\cos(6x)\cos(3x)}
\]

Factor numerator and denominator:
\[
= \frac{2\sin(6x)(\cos x + \cos 3x)}{2\cos(6x)(\cos x + \cos 3x)}
\]

Cancel common terms:
\[
= \frac{\sin(6x)}{\cos(6x)} = \tan(6x)
\]

Therefore,
\[
\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x \qquad \textbf{Hence Proved.}
\]