Prove that: $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$

We use the trigonometric identities: $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$ and $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$.
$L.H.S. = \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}$
Applying the identities to the numerator and denominator:
$= \frac{2 \sin \left( \frac{7x+5x}{2} \right) \cos \left( \frac{7x-5x}{2} \right) + 2 \sin \left( \frac{9x+3x}{2} \right) \cos \left( \frac{9x-3x}{2} \right)}{2 \cos \left( \frac{7x+5x}{2} \right) \cos \left( \frac{7x-5x}{2} \right) + 2 \cos \left( \frac{9x+3x}{2} \right) \cos \left( \frac{9x-3x}{2} \right)}$
$= \frac{2 \sin 6x \cos x + 2 \sin 6x \cos 3x}{2 \cos 6x \cos x + 2 \cos 6x \cos 3x}$
$= \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)}$
$= \frac{\sin 6x}{\cos 6x} = \tan 6x = R.H.S.$