Example: Find the value of $\arcsin \frac{1}{\sqrt{10}}+\arccos \frac{5}{\sqrt{26}}$ $+\operatorname{arctg} \frac{1}{7}+\arccos \frac{8}{\sqrt{65}}$.
Let the four terms on the left side of the equation be denoted as $\alpha, \beta, \gamma, \delta$, clearly they all belong to $\left(0, \frac{\pi}{4}\right)$.

From $\sin \alpha=\frac{1}{\sqrt{10}}$ and $\alpha \in\left(0, \frac{\pi}{4}\right)$, it is easy to get $\operatorname{tg} \alpha = \frac{1}{3}, a=\operatorname{arctg} \frac{1}{3}$. Also, when $a>0, b>0$, $\operatorname{arctg} \frac{b}{a}=\arg (a+b i)$, so $\alpha=\arg (3+i)$.
Similarly, $\beta=\arg (5+i), \gamma=\arg (7+i)$, $\delta=\arg (8+i)$.

Thus, $\alpha+\beta+\gamma+\delta=\arg (3+i) + \arg (5+i) + \arg (7+i) + \arg (8+i) \in \operatorname{Arg}[(3+i)(5+i)(7+i)(8+i)] = \operatorname{Arg}[650(1+i)] = \operatorname{Arg}(1+i) = \left\{\frac{\pi}{4}+2 k \pi, \quad k \in \mathbb{Z}\right\}$.
Since $0<\alpha+\beta+\gamma+\delta<\pi$.
Therefore, take $k=0$.
Thus, $\alpha+\beta+\gamma+\delta=\frac{\pi}{4}=$ the original expression.