Answer
See the explanation below.
Work Step by Step
One of the sum-to-product formulas is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. Therefore, $\cos 4x-\cos 2x$ can be written as given below:
$\cos 4x-\cos 2x=-2\sin \frac{4x+2x}{2}\sin \frac{4x-2x}{2}$
One of the sum-to-product formulas is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. Therefore, $\sin 2x-\sin 4x$ can be written as given below:
$\sin 2x-\sin 4x=2\sin \frac{2x-4x}{2}\cos \frac{2x+4x}{2}$
Now, consider the left side of the provided expression:
$\frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}$
The expression can be simplified as provided below:
$\begin{align}
& \frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\frac{-2\sin \frac{4x+2x}{2}\sin \frac{4x-2x}{2}}{2\sin \frac{2x-4x}{2}\cos \frac{2x+4x}{2}} \\
& =\frac{-2\sin \left( \frac{6x}{2} \right)\sin \left( \frac{2x}{2} \right)}{2\sin \left( \frac{-2x}{2} \right)\cos \left( \frac{6x}{2} \right)} \\
& =\frac{-2\sin 3x\sin x}{2\sin \left( -x \right)\cos 3x}
\end{align}$
One of the even-odd identities of trigonometry is $\sin \left( -x \right)=-\sin x$. Therefore, applying this identity $2\sin \left( -x \right)$ can be written as $-2\sin x$. Then, the expression can be further simplified as:
$\begin{align}
& \frac{-2\sin 3x\sin x}{2\sin \left( -x \right)\cos 3x}=\frac{-2\sin 3x\sin x}{-2\sin x\cos 3x} \\
& =\frac{\sin 3x}{\cos 3x}
\end{align}$
Now, using one of the quotient identities of trigonometry, which is $\tan x=\frac{\sin x}{\cos x}$, the expression can be further written as:
$\frac{\sin 3x}{\cos 3x}=\tan 3x$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\tan 3x$
