L.H.S. $= \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}$

Using Identity 20(iii), $\sin x + \sin y = 2\sin \frac{x+y}{2}\cos \frac{x-y}{2}$, we simplify the numerator terms:

$$\sin 7x + \sin 5x = 2\sin 6x \cos x$$

and

$$\sin 9x + \sin 3x = 2\sin 6x \cos 3x$$

Therefore, the Numerator $= 2\sin 6x \cos x + 2\sin 6x \cos 3x = 2\sin 6x (\cos x + \cos 3x)$.

Using Identity 20(i), $\cos x + \cos y = 2\cos \frac{x+y}{2}\cos \frac{x-y}{2}$, we simplify the denominator terms:

$$\cos 7x + \cos 5x = 2\cos 6x \cos x$$

and

$$\cos 9x + \cos 3x = 2\cos 6x \cos 3x$$

Therefore, the Denominator $= 2\cos 6x \cos x + 2\cos 6x \cos 3x = 2\cos 6x (\cos x + \cos 3x)$.

Substituting these simplified forms back into the original expression:

$$\text{L.H.S.} = \frac{2\sin 6x (\cos x + \cos 3x)}{2\cos 6x (\cos x + \cos 3x)}$$

By cancelling the common factor $2(\cos x + \cos 3x)$ from the numerator and denominator, we get:

$$\text{L.H.S.} = \frac{\sin 6x}{\cos 6x} = \tan 6x = \text{R.H.S.}$$

Hence proved.