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, $\sin 3x+\sin x$ can be written as given below:
$\sin 3x+\sin x=2\sin \frac{3x+x}{2}\cos \frac{3x-x}{2}$
Another sum-to-product formula is $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. Thus, $\cos 3x+\cos x$ can be written as given below:
$\cos 3x+\cos x=2\cos \frac{3x+x}{2}\cos \frac{3x-x}{2}$
Now, consider the left-hand side of the provided expression:
$\frac{\sin 2x+\left( \sin 3x+\sin x \right)}{\cos 2x+\left( \cos 3x+\cos x \right)}$
And the expression can be simplified as shown below:
$\begin{align}
& \frac{\sin 2x+\left( \sin 3x+\sin x \right)}{\cos 2x+\left( \cos 3x+\cos x \right)}=\frac{\sin 2x+2\sin \left( \frac{3x+x}{2} \right)\cos \left( \frac{3x-x}{2} \right)}{\cos 2x+2\cos \left( \frac{3x+x}{2} \right)\cos \left( \frac{3x-x}{2} \right)} \\
& =\frac{\sin 2x+2\sin \left( \frac{4x}{2} \right)\cos \left( \frac{2x}{2} \right)}{\cos 2x+2\cos \left( \frac{4x}{2} \right)\cos \left( \frac{2x}{2} \right)} \\
& =\frac{\sin 2x+2\sin 2x\cos x}{\cos 2x+2\cos 2x\cos x}
\end{align}$
Then, by taking out the common term, that is $\sin 2x$, in the numerator and $\cos 2x$ in the denominator, we get:
$\begin{align}
& \frac{\sin 2x+2\sin 2x\cos x}{\cos 2x+2\cos 2x\cos x}=\frac{\sin 2x\left( 1+2\cos x \right)}{\cos 2x\left( 1+2\cos x \right)} \\
& =\frac{\sin 2x}{\cos 2x}
\end{align}$
Now, using one of the quotient identities of trigonometry, $\tan x=\frac{\sin x}{\cos x}$, the expression can be further written as:
$\frac{\sin 2x}{\cos 2x}=\tan 2x$
Thus, the left-hand side of the expression is equal to the right-hand side:
$\frac{\sin 2x+\left( \sin 3x+\sin x \right)}{\cos 2x+\left( \cos 3x+\cos x \right)}=\tan 2x$.
