3.140. $\frac{\sin 13 \alpha+\sin 14 \alpha+\sin 15 \alpha+\sin 16 \alpha}{\cos 13 \alpha+\cos 14 \alpha+\cos 15 \alpha+\cos 16 \alpha}$.
Solution.

$$
\begin{aligned}
& \frac{\sin 13 \alpha+\sin 14 \alpha+\sin 15 \alpha+\sin 16 \alpha}{\cos 13 \alpha+\cos 14 \alpha+\cos 15 \alpha+\cos 16 \alpha}= \\
& =\frac{(\sin 16 \alpha+\sin 13 \alpha)+(\sin 15 \alpha+\sin 14 \alpha)}{(\cos 16 \alpha+\cos 13 \alpha)+(\cos 15 \alpha+\cos 14 \alpha)}= \\
& =\frac{2 \sin \frac{29 \alpha}{2} \cos \frac{3}{2} \alpha+2 \sin \frac{29 \alpha}{2} \cos \frac{\alpha}{2}}{2 \cos \frac{29 \alpha}{2} \cos \frac{3}{2} \alpha+2 \cos \frac{29 \alpha}{2} \cos \frac{\alpha}{2}}=\frac{2 \sin \frac{29 \alpha}{2}\left(\cos \frac{3}{2} \alpha+\cos \frac{\alpha}{2}\right)}{2 \cos \frac{29 \alpha}{2}\left(\cos \frac{3}{2} \alpha+\cos \frac{\alpha}{2}\right)}= \\
& =\frac{2 \sin \frac{29 \alpha}{2}}{2 \cos \frac{29 \alpha}{2}}=\tan \frac{29 \alpha}{2}
\end{aligned}
$$

Answer: $\tan \frac{29 \alpha}{2}$.