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

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

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

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