8. If $(x+1)(y+1)=2$, then $\arctan x+\arctan y=$
8. $\frac{\pi}{4}$ or $-\frac{3 \pi}{4}$.
$$
\left.\begin{array}{l}
\tan (\arctan x+\arctan y)=\frac{x+y}{1-x y} \\
(x+1)(y+1)=2 \Rightarrow(x+y)+x y=1
\end{array}\right\} \Rightarrow \tan (\arctan x+\arctan y)=1 .
$$

Since $-\pi<\arctan x+\arctan y<\pi$, it follows that $\arctan x+\arctan y=-\frac{3 \pi}{4}$ or $\frac{\pi}{4}$.