Prove that $\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$.

We have $L.H.S. = \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x}$.
Using the sum-to-product formula $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$ and $\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$:
$L.H.S. = \frac{(\sin 5x + \sin x) - 2\sin 3x}{\cos 5x - \cos x}$
$L.H.S. = \frac{2\sin 3x \cos 2x - 2\sin 3x}{-2\sin 3x \sin 2x}$
$L.H.S. = \frac{2\sin 3x(\cos 2x - 1)}{-2\sin 3x \sin 2x}$
$L.H.S. = -\frac{\cos 2x - 1}{\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$
Using double angle identities $1 - \cos 2x = 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:
$L.H.S. = \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x = R.H.S.$