Proves of identities in inverse trigonometry Can someone please help me prove the following results from inverse trigonometry?
$$\tan^{-1}x + \tan^{-1}y  = \pi + \tan^{-1}\frac{x+y}{1-xy}( x>0, y>0, xy>1)$$
and
$$\tan^{-1}x + \tan^{-1}y  = -\pi + \tan^{-1}\frac{x+y}{1-xy} ( x<0, y< 0, xy>1)$$
I know the prove for $\tan^{-1}x + \tan^{-1}y  =\tan^{-1}\frac{x+y}{1-xy} (  xy<1)$ but cant prove the other two. Please do help.
Thanks in advance :)
 A: We have
$$\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}$$
so let 
$$X=\arctan x\quad;\quad Y=\arctan y$$
hence
$$\tan(\arctan x+\arctan y)=\frac{x+y}{1-xy}$$
Now it's easy to prove that
$$\arctan x+\arctan \frac 1 x=\frac \pi 2,\quad x>0$$
so if 
$$xy>1\iff y>\frac 1 x$$
$$\arctan x+\arctan y\in(\frac \pi 2,\pi)$$
since $$y>0\iff \arctan  y\in(0,\frac \pi 2)$$
and if 
$$z\in (\frac \pi 2,\pi)\iff z-\pi\in(-\frac\pi 2,0)$$
then
$$\arctan(\tan (z-\pi))=z-\pi=\arctan(\tan z)$$ 
finaly
$$\arctan x+\arctan y-\pi=\arctan(\tan(\arctan x+\arctan y))=\arctan\left(\frac{x+y}{1-xy}\right)$$
A: Apply $\;\tan\;$ to both sides and use trigonometric identities:
$$\begin{align*}\bullet&\tan (\arctan x+\arctan y)=\frac{\tan\arctan x+\tan\arctan y}{1-\tan\arctan x\tan\arctan y}=\frac{x+y}{1-xy}\\{}\\\bullet&\tan\left(\pi+\arctan\frac{x+y}{1-xy}\right)=\tan\arctan\frac{x+y}{1-xy}=\frac{x+y}{1-xy}\end{align*}$$
A: Hint:  Let $\tan^{-1} x = a$ and $\tan^{-1} y = b$. Then $x = \tan a$ and $y = \tan b$.
$$\frac{x+y}{1-xy} = \frac{\tan a + \tan b}{1- \tan a \tan b}$$
$$\frac{x+y}{1-xy} = \tan {(a+b)}$$
