1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+\operatorname{tg} \alpha+\frac{5}{2} \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=-\frac{1}{2}$. Find $\operatorname{ctg} \alpha$.
Answer: $\frac{1}{3}$ or $-\frac{6}{7}$.

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

$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \operatorname{tg} 2 \gamma}+\operatorname{tg} \alpha+\frac{5}{2} \operatorname{tg} 2 \gamma=0 \Leftrightarrow \frac{\operatorname{tg} \alpha-\frac{4}{3}}{1+\frac{4}{3} \operatorname{tg} \alpha}+\operatorname{tg} \alpha-\frac{10}{3}=0
$$

By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $6 \operatorname{tg}^{2} \alpha-11 \operatorname{tg} \alpha-21=0$, from which $\operatorname{tg} \alpha=3$ or $\operatorname{tg} \alpha=-\frac{7}{6}$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \alpha=\frac{1}{3}$ or $\operatorname{ctg} \alpha=-\frac{6}{7}$.