3.134. $\frac{\sin 4 \alpha+\sin 5 \alpha+\sin 6 \alpha}{\cos 4 \alpha+\cos 5 \alpha+\cos 6 \alpha}$.
Solution.

$\frac{\sin 4 \alpha+\sin 5 \alpha+\sin 6 \alpha}{\cos 4 \alpha+\cos 5 \alpha+\cos 6 \alpha}=\frac{(\sin 4 \alpha+\sin 6 \alpha)+\sin 5 \alpha}{(\cos 4 \alpha+\cos 6 \alpha)+\cos 5 \alpha}=$

$=\frac{2 \sin 5 \alpha \cos \alpha+\sin 5 \alpha}{2 \cos 5 \alpha \cos \alpha+\cos 5 \alpha}=\frac{\sin 5 \alpha(2 \cos \alpha+1)}{\cos 5 \alpha(2 \cos \alpha+1)}=\frac{\sin 5 \alpha}{\cos 5 \alpha}=\tan 5 \alpha$.

Answer: $\tan 5 \alpha$.