8.401. $\left\{\begin{array}{l}\tan \frac{x}{2}+\tan \frac{y}{2}=2 \\ \cot x+\cot y=-1.8 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \sin y \neq 0 .\end{array}\right.$

Using the formula $\operatorname{ctg} \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{2 \operatorname{tg} \frac{\alpha}{2}}$, we can write the system of equations as

$$
\begin{aligned}
& \left\{\begin{array}{l}
\operatorname{tg} \frac{y}{2}=2-\operatorname{tg} \frac{x}{2} \\
\frac{1-\operatorname{tg}^{2} \frac{x}{2}}{2 \operatorname{tg} \frac{x}{2}}+\frac{1-\operatorname{tg}^{2} \frac{y}{2}}{2 \operatorname{tg} \frac{y}{2}}=-1.8
\end{array} \Rightarrow \operatorname{tg} \frac{y}{2}=2-\operatorname{tg} \frac{x}{2},\right. \\
& \frac{1-\operatorname{tg}^{2} \frac{x}{2}}{2 \operatorname{tg} \frac{x}{2}}+\frac{1-\left(2-\operatorname{tg}^{2} \frac{x}{2}\right)^{2}}{2\left(2-\operatorname{tg} \frac{x}{2}\right)}=-1.8 ; 4 \operatorname{tg}^{2} \frac{x}{2}-8 \operatorname{tg} \frac{x}{2}-5=0 .
\end{aligned}
$$

Solving this equation as a quadratic equation in $\operatorname{tg} \frac{x}{2}$, we get $\operatorname{tg} \frac{x}{2}=-\frac{1}{2}, \operatorname{tg} \frac{x}{2}=\frac{5}{2} \cdot$ Further, we have

1) $\left\{\begin{array}{l}\operatorname{tg} \frac{y_{1}}{2}=\frac{5}{2}, \\ \operatorname{tg} \frac{x_{1}}{2}=-\frac{1}{2} \text { or 2) }\end{array}\left\{\begin{array}{l}\operatorname{tg} \frac{y_{2}}{2}=-\frac{1}{2} \\ \operatorname{tg} \frac{x_{2}}{2}=\frac{5}{2}\end{array}\right.\right.$
2) $\left\{\begin{array}{l}\frac{x_{1}}{2}=-\operatorname{arctg} \frac{1}{2}+\pi k_{1}, \\ \frac{y_{1}}{2}=\operatorname{arctg} \frac{5}{2}+\pi k_{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{1}, \\ y_{1}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{2} .\end{array}\right.\right.$
3) $\left\{\begin{array}{l}\frac{x_{2}}{2}=\operatorname{arctg} \frac{5}{2}+\pi k_{1}, \\ \frac{y_{2}}{2}=-\operatorname{arctg} \frac{1}{2}+\pi k_{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{2}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{1}, \\ y_{2}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{2},\end{array}\right.\right.$ where $k_{1}$ and $k_{2} \in Z$.

Answer: $x_{1}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{1}, y_{1}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{2}$;

$$
x_{2}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{1}, y_{2}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{2}, \text { where } k_{1} \text { and } k_{2} \in Z
$$