Simplify the expression (cos 6B + cos 2B) / (sin 6B + sin 2B) using sum-to-product identities.

The relevant identities are:

$$\cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$$ $$\sin X + \sin Y = 2 \sin \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$$

For the numerator, we have $\cos 6B + \cos 2B$. Let $X = 6B$ and $Y = 2B$.

$$\cos 6B + \cos 2B = 2 \cos \left(\frac{6B+2B}{2}\right) \cos \left(\frac{6B-2B}{2}\right)$$ $$= 2 \cos \left(\frac{8B}{2}\right) \cos \left(\frac{4B}{2}\right)$$ $$= 2 \cos(4B) \cos(2B)$$

For the denominator, we have $\sin 6B + \sin 2B$. Let $X = 6B$ and $Y = 2B$.

$$\sin 6B + \sin 2B = 2 \sin \left(\frac{6B+2B}{2}\right) \cos \left(\frac{6B-2B}{2}\right)$$ $$= 2 \sin \left(\frac{8B}{2}\right) \cos \left(\frac{4B}{2}\right)$$ $$= 2 \sin(4B) \cos(2B)$$

Cancel out the common terms 2 and $\cos(2B)$ (assuming $\cos(2B) \neq 0$).

$$= \frac{\cos(4B)}{\sin(4B)}$$

Using the identity $\frac{\cos \theta}{\sin \theta} = \cot \theta$:

$$= \cot(4B)$$

The simplified expression is $\cot(4B)$.