$3.431 \sin \left(2 \operatorname{arctg} \frac{1}{2}\right)+\operatorname{tg}\left(\frac{1}{2} \arcsin \frac{15}{17}\right)=\frac{7}{5}$. 

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$3.431 \sin \left(2 \operatorname{arctan} \frac{1}{2}\right)+\operatorname{tan}\left(\frac{1}{2} \arcsin \frac{15}{17}\right)=\frac{7}{5}$.
## Solution.

$$
\begin{aligned}
& \text { Let } \alpha=\operatorname{arctg} \frac{1}{2}, \beta=\arcsin \frac{15}{17}, \text { then } \operatorname{tg} \alpha=\frac{1}{2}, 0<\alpha<\frac{\pi}{2}, \\
& \sin \beta=\frac{15}{17}, \cos \beta=\frac{8}{17} \Rightarrow \sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{4}{5}, \operatorname{tg} \frac{\beta}{2}=\frac{\sin \beta}{1+\cos \beta}=\frac{3}{5} \Rightarrow \\
& \Rightarrow \sin 2 \alpha+\operatorname{tg} \frac{\beta}{2}=\frac{7}{5} . \\
& \quad \cdot \\
& 3.432 \sin \left(2 \operatorname{arctg} \frac{1}{2}\right)-\operatorname{tg}\left(\frac{1}{2} \arcsin \frac{15}{17}\right)=\frac{1}{5} .
\end{aligned}
$$

Solution. This problem is analogous to problem 3.431.