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