$\frac{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}{\operatorname{ctg} \alpha-\operatorname{ctg} \beta}+\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=?$
$$
\begin{aligned}
\frac{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}{\operatorname{ctg} \alpha-\operatorname{ctg} \beta} & =\frac{\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}}{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos \beta}{\sin \beta}}=\frac{\cos \alpha \sin \beta+\sin \alpha \cos \beta}{\cos \alpha \sin \beta-\sin \alpha \cos \beta}= \\
& =\frac{\sin (\alpha+\beta)}{-\sin (\alpha-\beta)}=-\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}
\end{aligned}
$$

thus

$$
\frac{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}{\operatorname{ctg} \alpha-\operatorname{ctg} \beta}+\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=0
$$

(Leó Spitzer, Budapest, V. district.)

Number of solutions: 18.