Prove $\arcsin \frac{1}{10}=\operatorname{arctg} \frac{1}{3}=\arg (3+i)$
$$\begin{array}{l} \arccos \frac{5}{\sqrt{26}}=\operatorname{arctg} \frac{1}{5} \\ =\arg (5+i), \\ \operatorname{arctg} \frac{1}{7}=\arg (7+i), \\ \operatorname{arcctg} 8=\arg (8+i) . \\ \therefore \quad \text { the left side of the original equation }=\arg (3+i)+\arg (5 \\ +i)+\arg (7+i)+\arg (8+i) \\ \in \operatorname{Arg}[(3+i)(5+i)(7+i) \\ (8+i)]=\operatorname{Arg}(650+650 i) . \end{array}$$

But $0<\operatorname{arcctg} 8<\operatorname{arctg} \frac{1}{7}<\operatorname{arctg} \frac{1}{5}$
$<\operatorname{arctg} \frac{1}{3}<\pi$,
the left side of the original equation $\in(0, \pi)$, while Arg（650 $+650 i)=n \pi+\frac{\pi}{4}, n \in Z$.
$n$ should be 0, so the left side of the original equation $=\frac{\pi}{4}$.