3.110. $\frac{\cos 3 \alpha+\cos 4 \alpha+\cos 5 \alpha}{\sin 3 \alpha+\sin 4 \alpha+\sin 5 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{\cos 3 \alpha + \cos 4 \alpha + \cos 5 \alpha}{\sin 3 \alpha + \sin 4 \alpha + \sin 5 \alpha} = \frac{2 \cos 4 \alpha \cos \alpha + \cos 4 \alpha}{2 \sin 4 \alpha \cos \alpha + \sin 4 \alpha} = \frac{\cos 4 \alpha (2 \cos \alpha + 1)}{\sin 4 \alpha (2 \cos \alpha + 1)} = \\
& = \frac{\cos 4 \alpha}{\sin 4 \alpha} = \operatorname{ctg} 4 \alpha
\end{aligned}
$$

Answer: $\operatorname{ctg} 4 \alpha$.