Simplify the trigonometric expression: (sin 3θ + sin θ) / (cos 3θ + cos θ)

The sum-to-product identities are:
$\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

For the numerator, let $A = 3\theta$ and $B = \theta$:

$$\sin 3\theta + \sin \theta = 2 \sin \left(\frac{3\theta+\theta}{2}\right) \cos \left(\frac{3\theta-\theta}{2}\right)$$ $$= 2 \sin \left(\frac{4\theta}{2}\right) \cos \left(\frac{2\theta}{2}\right)$$ $$= 2 \sin 2\theta \cos \theta$$

For the denominator, let $A = 3\theta$ and $B = \theta$:

$$\cos 3\theta + \cos \theta = 2 \cos \left(\frac{3\theta+\theta}{2}\right) \cos \left(\frac{3\theta-\theta}{2}\right)$$ $$= 2 \cos \left(\frac{4\theta}{2}\right) \cos \left(\frac{2\theta}{2}\right)$$ $$= 2 \cos 2\theta \cos \theta$$

Cancel out the common terms 2 and $\cos \theta$ (assuming $\cos \theta \neq 0$):

$$\frac{2 \sin 2\theta \cos \theta}{2 \cos 2\theta \cos \theta} = \frac{\sin 2\theta}{\cos 2\theta}$$

$$\frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta$$

Thus, we have shown that $\frac{\sin 3\theta + \sin \theta}{\cos 3\theta + \cos \theta} = \tan 2\theta$.

The final answer is $\text{Shown}$.