1. It is known that $\operatorname{tg}(2 \alpha-\beta)+6 \operatorname{tg} 2 \alpha+\operatorname{tg} \beta=0, \operatorname{tg} \alpha=2$. Find $\operatorname{ctg} \beta$.
Answer: 1 or $\frac{1}{7}$.

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

$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorname{tg} \beta}+6 \operatorname{tg} 2 \alpha+\operatorname{tg} \beta=0 \Leftrightarrow \frac{-\frac{4}{3}-\operatorname{tg} \beta}{1-\frac{4}{3} \operatorname{tg} \beta}-8+\operatorname{tg} \beta=0
$$

By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $\operatorname{tg}^{2} \beta-8 \operatorname{tg} \beta+7=0$, from which $\operatorname{tg} \beta=7$ or $\operatorname{tg} \beta=1$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \beta=\frac{1}{7}$ or $\operatorname{ctg} \beta=1$.