3.137. $\frac{\cos 7 \alpha-\cos 8 \alpha-\cos 9 \alpha+\cos 10 \alpha}{\sin 7 \alpha-\sin 8 \alpha-\sin 9 \alpha+\sin 10 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{\cos 7 \alpha - \cos 8 \alpha - \cos 9 \alpha + \cos 10 \alpha}{\sin 7 \alpha - \sin 8 \alpha - \sin 9 \alpha + \sin 10 \alpha} = \frac{(\cos 10 \alpha + \cos 7 \alpha) - (\cos 9 \alpha + \cos 8 \alpha)}{(\sin 10 \alpha + \sin 7 \alpha) - (\sin 9 \alpha + \sin 8 \alpha)} = \\
& = \frac{2 \cos \frac{17 \alpha}{2} \cos \frac{3 \alpha}{2} - 2 \cos \frac{17 \alpha}{2} \cos \frac{\alpha}{2}}{2 \sin \frac{17 \alpha}{2} \cos \frac{3 \alpha}{2} - 2 \sin \frac{17 \alpha}{2} \cos \frac{\alpha}{2}} = \frac{2 \cos \frac{17 \alpha}{2} \left( \cos \frac{3 \alpha}{2} - \cos \frac{\alpha}{2} \right)}{2 \sin \frac{17 \alpha}{2} \left( \cos \frac{3 \alpha}{2} - \cos \frac{\alpha}{2} \right)} = \\
& = \frac{\cos \frac{17 \alpha}{2}}{\sin \frac{17 \alpha}{2}} = \operatorname{ctg} \frac{17 \alpha}{2}
\end{aligned}
$$

Answer: $\operatorname{ctg} \frac{17 \alpha}{2}$.