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!

I think this might be an idea for you.
Let  $$f(x,y)= \arctan(x)+\arctan(y)-\arctan(\frac{x+y}{1-xy})$$ defined for  $xy> 1$, then we have 
$$ \frac{ \partial f}{\partial x }= \frac{1}{1+x^2} -\frac{(1-xy)+y(x+y)}{(1-xy)^2 +(x+y)^2 }= \frac{1}{1+x^2} -\frac{1+y^2}{(1+y^2)(1+x^2) }=0 $$
Similarly you can show that $ \frac{ \partial f}{\partial y}=0$. Hence our function is constant, i.e. $f(x,y)=C$. Now let $x=\tan(\frac{\pi}{3})= \sqrt{3}$ and  $y=\tan(\frac{\pi}{4})=1$  then $xy \approx  1.732050807568877 > 1$. On the other hand  $$f(x,y)=\frac{\pi}{3} +\frac{\pi}{4}- arctan \Big( \frac{ 1+\sqrt{3}}{1-\sqrt{3}} \Big) =\frac{\pi}{3} +\frac{\pi}{4} -(-\frac{5\pi}{12})= \pi  $$ However  $f(x,y)$ is constant, and so  $f(x,y)= \pi$ for all $xy>1$.
