11.1. Angles $\alpha$ and $\beta$ are such that $\operatorname{tg} \alpha+\operatorname{tg} \beta=2$, and $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=5$. Find the value of $\operatorname{tg}(\alpha+\beta)$.
Answer: $10 / 3$.

Solution: Transform the condition $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=5: \frac{1}{\operatorname{tg} \alpha}+\frac{1}{\operatorname{tg} \beta}=5$, which is $\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=5$, or $\frac{2}{\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=5$, from which $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{2}{5}$. Finally, $\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=\frac{2}{1-\frac{2}{5}}=\frac{2 \cdot 5}{5-2}=\frac{10}{3}$.

Criteria: Only the answer - 0 points. Found $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{2}{5}$ without further substantial progress (no further solution or with an error) - 3 points. 7 points for a complete and correct solution.