Problem 6. Calculate $2 \operatorname{arctg} 5+\arcsin \frac{5}{13}$.
Answer: $\pi$.

Solution. First solution. Let $\operatorname{arctg} 5$ be denoted by $\alpha$, and $\arcsin 5 / 13$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=5^{2} / 12^{2}$, from which $\operatorname{tg} \beta=5 / 12$; also $\operatorname{tg} \alpha=5, \alpha \in(0, \pi / 2)$.

We find:

$$
\begin{gathered}
\operatorname{tg}(2 \alpha)=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{2 \cdot 5}{1-5^{2}}=-\frac{10}{24}=-\frac{5}{12} \\
\operatorname{tg}(2 \alpha+\beta)=\frac{\operatorname{tg}(2 \alpha)+\operatorname{tg} \beta}{1-\operatorname{tg}(2 \alpha) \operatorname{tg} \beta}=\frac{-5 / 12+5 / 12}{1-(-5 / 12) \cdot(5 / 12)}=0
\end{gathered}
$$

Finally, since $0<\alpha<\pi / 2,0<\beta<\pi / 2$, then $0<2 \alpha+\beta<3 \pi / 2$. Therefore, $2 \alpha+\beta=\pi$.