1. It is known that $\operatorname{tg}(2 \alpha-\beta)-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0, \operatorname{tg} \alpha=-3$. Find $\operatorname{ctg} \beta$.
Answer: -1 or $\frac{4}{3}$.

Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{-6}{1-9}=\frac{3}{4}$. Then the given equality can be transformed as follows:

$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorname{tg} \beta}-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0 \Leftrightarrow \frac{\frac{3}{4}-\operatorname{tg} \beta}{1+\frac{3}{4} \operatorname{tg} \beta}-3+4 \operatorname{tg} \beta=0
$$

By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $4 \operatorname{tg}^{2} \beta+\operatorname{tg} \beta-3=0$, from which $\operatorname{tg} \beta=\frac{3}{4}$ or $\operatorname{tg} \beta=-1$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \beta=\frac{4}{3}$ or $\operatorname{ctg} \beta=-1$.