TITLE: What is $\arctan(x) + \arctan(y)$
QUESTION [17 upvotes]: I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex] 
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex] 
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you.

REPLY [4 votes]: 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):
$$
x>0, y>0 \rightarrow t<0 \rightarrow \gamma<0\rightarrow \alpha+\beta=\gamma+\pi
$$
$$
x<0, y<0 \rightarrow t>0 \rightarrow \gamma>0\rightarrow \alpha+\beta=\gamma-\pi
$$