Prove that
$\dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$
$\sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha -\beta}{2}\right)$
$\cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)$
$\begin{aligned}
\text{LHS} & = \dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} \\
& \\
& = \dfrac{2 \sin \left(\dfrac{5x+x}{2}\right) \cos \left(\dfrac{5x-x}{2}\right)-2\sin 3x}{-2\sin \left(\dfrac{5x+x}{2}\right) \sin \left(\dfrac{5x-x}{2}\right)} \\
& \\
& = \dfrac{2 \sin 3x \cos 2x - 2 \sin 3x}{-2 \sin 3x \cdot \sin 2x} \\
& \\
& = \dfrac{2 \sin 3x \left(\cos 2x - 1\right)}{-2\sin 3x \cdot \sin 2x} \\
& \\
& = \dfrac{\cos 2x - 1}{-\sin 2x} \\
& \\
& = \dfrac{-2 \sin^2 x}{-2\sin x \cos x} \qquad \left[\text{Note: }\cos 2\theta - 1 = -2 \sin^2 \theta \; ; \quad \sin 2x = 2 \sin x \cos x\right] \\
& \\
& = \dfrac{\sin x}{\cos x} \\
& \\
& = \tan x = \text{RHS}
\end{aligned}$