$$
\begin{array}{ll}
1^{\circ} . & \frac{\sin 3 \alpha+\sin \alpha}{\cos 3 \alpha-\cos \alpha}+\operatorname{ctg} \alpha=? \\
2^{\circ} . & \frac{\sin 3 \alpha-\sin \alpha}{\cos 3 \alpha+\cos \alpha}-\operatorname{tg} \alpha=?
\end{array}
$$


$1^{\circ} . \quad \frac{\sin 3 \alpha+\sin \alpha}{\cos 3 \alpha-\cos \alpha}+\operatorname{ctg} \alpha=\frac{2 \sin \frac{3 \alpha+\alpha}{2} \cos \frac{3 \alpha-\alpha}{2}}{-2 \sin \frac{3 \alpha+\alpha}{2} \sin \frac{3 \alpha-\alpha}{2}}+\operatorname{ctg} \alpha=-\frac{\cos \alpha}{\sin \alpha}+\operatorname{ctg} \alpha=0$

$2^{\circ} . \quad \frac{\sin 3 \alpha-\sin \alpha}{\cos 3 \alpha+\cos \alpha}-\operatorname{tg} \alpha=\frac{2 \cos \frac{3 \alpha+\alpha}{2} \sin \frac{3 \alpha-\alpha}{2}}{2 \cos \frac{3 \alpha+\alpha}{2} \cos n \frac{3 \alpha-\alpha}{2}}-\operatorname{tg} \alpha=\frac{\sin \alpha}{\cos \alpha}-\operatorname{tg} \alpha=0$.

(Neumann Frida, Budapest.)

