3.105. $\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}$.
## Solution.

$\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (\pi-6 \alpha)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos 6 \alpha}{\cos 2 \alpha}=$
$=\frac{\sin 6 \alpha \cos 2 \alpha-\cos 6 \alpha \sin 2 \alpha}{\sin 2 \alpha \cos 2 \alpha}=\frac{\sin 4 \alpha}{\sin 2 \alpha \cos 2 \alpha}=\frac{2 \sin 4 \alpha}{2 \sin 2 \alpha \cos 2 \alpha}=$

$=\frac{2 \sin 4 \alpha}{\sin 4 \alpha}=2$.

Answer: 2.