Example 8. Find the value of $2 \operatorname{arctg} \frac{1}{5}+\operatorname{arctg} \frac{1}{4}$.
This problem is to find the inverse function, but using the tangent makes the calculation easier.

Slightly solving, let $\operatorname{arctg} \frac{1}{5}=\alpha$, $\operatorname{arc} \operatorname{tg} \frac{1}{4}=\beta$, then $\operatorname{tg} \alpha=\frac{1}{5}, \operatorname{tg} \beta=\frac{1}{4}$.
$\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{5}{12}$.
$$
\begin{array}{l}
\operatorname{tg}\left(2 \operatorname{arctg} \frac{1}{5}+\operatorname{arctg} \frac{1}{4}\right) \\
=\operatorname{tg}(2 \alpha+\beta) \\
=\frac{\operatorname{tg} 2 \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} 2 \alpha \cdot \operatorname{tg} \beta}=\frac{32}{43} .
\end{array}
$$

From the range of $2 \alpha+\beta$, it is easy to get
$2 \operatorname{arctg} \frac{1}{5}+\operatorname{arctg} \frac{1}{4}=\operatorname{arctg} \frac{32}{43}$.