Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$

I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\frac{x+y}{1-xy}=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan(a+b)$
$\implies a+b=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b} $
$\implies \arctan x+\arctan y = \arctan\frac{x+y}{1-xy}$
I know $\pi$ is the period of $\tan x$ so $xy\gt 1$ constraint must have something to do with this. But I am not able to figure out how exactly these period and $xy\gt 1$ are related. Any help is appreciated. Thanks!

Here is an insight given by a purely geometrical proof. Consider the Figure below.

Supposing $x>0$, the right-angled triangle $\triangle ABC$ has sides $\overline{AB} = 1$ and $\overline{BC} = x$, so that
$$\alpha = \arctan x.$$
Extend side $AB$ with a segment $\overline{BD}=\frac{x}{y}$, giving
$$\gamma = \arctan y.$$
Now draw from $D$ the line segment $DH$ perpendicular to $AC$ and note that $\triangle ADH \sim \triangle ABC$, with a scaling factor given by 
$$\frac{\overline{AD}}{\overline{AC}} = \frac{x+y}{y\sqrt{1+x^2}}$$
yielding
$$\overline{DH} =\frac{x(x+y)}{y\sqrt{1+x^2}},$$
$$\overline{AH} = \frac{x+y}{y\sqrt{1+x^2}},$$
and
\begin{eqnarray}
\overline{CH} &=&\overline{AC}-\overline{AH}=\\
&=&\sqrt{1+x^2}- \frac{x+y}{y\sqrt{1+x^2}}=\\
&=& \boxed{\frac{x(xy-1)}{y\sqrt{1+x^2}}}.\tag{1}\label{boxed}
\end{eqnarray}
By definition finally we have
\begin{eqnarray}
\beta &=& \arctan\left( \frac{\overline{DH}}{\overline{CH}}\right)=\\
&=& \arctan\left(\frac{x+y}{xy-1}\right).
\end{eqnarray}
Which, once you plug in definitions of $\alpha$, $\beta$, and $\gamma$ in 
\begin{equation}\alpha+\beta+\gamma=\pi,\end{equation}
leads to your identity. 
When $x<0$ it is sufficient to to define $\overline{BC} = -x$ and then use symmetries of the $\arctan(\cdot)$ function.

The constraint, from this perspective, comes from the boxed equation \eqref{boxed}. If $xy<1$, that would lead to a negative number. As a matter of fact, this is because in this case $\beta$ would be obtuse and the altitude $DH$ would be external, yielding a slightly different formula for $\overline{DH}$ and thus $\beta$. You can figure out which one as an excersise.
