What is $\arctan(x) + \arctan(y)$

Fix, as usual:

$$ -\frac{\pi}{2}<\gamma=\arctan(t)<\frac{\pi}{2} $$

now we have: $$ \tan (\gamma)=\tan(\alpha+\beta)=\frac{x+y}{1-xy}=t $$ and, if $xy>1$ we have the two cases ($x$ and $y$ have the same sign): $$ x>0, y>0 \rightarrow t<0 \rightarrow \gamma<0\rightarrow \alpha+\beta=\gamma+\pi $$ $$ x<0, y<0 \rightarrow t>0 \rightarrow \gamma>0\rightarrow \alpha+\beta=\gamma-\pi $$