4.83 It is known that $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{p}{q}$. Find $\operatorname{ctg} \beta$.
### 4.83 We have

$\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \frac{p}{q}$, or $\frac{\operatorname{ctg} \beta + \operatorname{ctg} \alpha}{\operatorname{ctg} \beta - \operatorname{ctg} \alpha} = \frac{p}{q}$.

Therefore, $\operatorname{ctg} \beta = \frac{p+q}{p-q} \operatorname{ctg} \alpha$.