Answer
See the explanation below.
Work Step by Step
One of the sum-to-product formulas is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. Thus, $\sin 3x-\sin x$ can be written as given below:
$\sin 3x-\sin x=2\sin \frac{3x-x}{2}\cos \frac{3x+x}{2}$
One of the sum-to-product formulas is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. Thus, $\cos 3x-\cos x$ can be written as given below:
$\cos 3x-\cos x=-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}$
Now, consider the left side of the given expression:
$\frac{\sin 3x-\sin x}{\cos 3x-\cos x}$
The expression can be simplified as given below:
$\begin{align}
& \frac{\sin 3x-\sin x}{\cos 3x-\cos x}=\frac{2\sin \left( \frac{3x-x}{2} \right)\cos \left( \frac{3x+x}{2} \right)}{-2\sin \left( \frac{3x+x}{2} \right)\sin \left( \frac{3x-x}{2} \right)} \\
& =\frac{2\sin \left( \frac{2x}{2} \right)\cos \left( \frac{4x}{2} \right)}{-2\sin \left( \frac{4x}{2} \right)\sin \left( \frac{2x}{2} \right)} \\
& =\frac{2\sin x\cos 2x}{-2\sin 2x\sin x} \\
& =-\frac{\cos 2x}{\sin 2x}
\end{align}$
Now, using one of the quotient identities of trigonometry, which is $\cot x=\frac{\cos x}{\sin x}$, the expression can be further written as:
$-\frac{\cos 2x}{\sin 2x}=-\cot 2x$
Therefore, the left side of the expression is equal to the right side, which is
$\frac{\sin 3x-\sin x}{\cos 3x-\cos x}=-\cot 2x$.
