proving $\tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$ (1) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$
(2) How ca we prove $\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$
$\bf{My\; Try::}$ My Try for $(1)\;,$ Let $\tan^{-1}(x)=A\Rightarrow x = \tan (A)$ and $\tan^{-1}(y) = B\Rightarrow y = \tan(B)$
Now $\displaystyle \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\cdot \tan B} = \frac{x+y}{1-xy}$
and Given $x>0\Rightarrow \tan A>0$ and $y>0\Rightarrow \tan B>0$ and given $xy>1$
So $\displaystyle \frac{x+y}{1-xy}<0\Rightarrow \tan (A+B)<0$
Means $(A+B)$ lies in $\bf(II)$ quadrant or in $\bf{(IV)}$ quadrant.
But above $\displaystyle \tan (A)\;\;,\tan (B)>0\Rightarrow 0<A,B<\frac{\pi}{2}$.
So $\displaystyle 0<A+B<\pi$(Means $(A+B)$ lies on $\bf{I}$ quadrant.)
Now I did not understand how can i solve it.
Help me
Thanks
 A: It is quite simple ;
lets consider tan(θ) = x and let it be so that tan(ϕ) = y , then we know , θ = tan⁻¹(x) and ϕ = tan⁻¹(y)
$\displaystyle\Rightarrow$ tan(θ+ϕ)= $\displaystyle(\frac{tanθ+tanϕ}{1-tanθ*tanϕ})\;$
$\displaystyle\Rightarrow$ tan(θ+ϕ) = $\displaystyle(\frac{x+y}{1-x*y})\;$ as [Note : tan(θ) = x and tan(ϕ) = y]
$\displaystyle\Rightarrow$ $\displaystyle \pi $ is an interval of tan , which means  tan(a-n*$\pi$) is same as tan(a) where $n∈ℤ$
using that :
$\displaystyle\Rightarrow$ tan(θ+ϕ-$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$
$\displaystyle\Rightarrow$ θ+ϕ-$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$
and alas simply :
∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$ when $x>0,y>0,xy>1$
conversly using n=-1 , we can observe ,
$\displaystyle\Rightarrow$ tan(θ+ϕ+$\displaystyle \pi)$ = $\displaystyle(\frac{x+y}{1-x*y})\;$
$\displaystyle\Rightarrow$ θ+ϕ+$\displaystyle \pi$ = tan⁻¹$\displaystyle(\frac{x+y}{1-x*y})\;$
hence,
∴$\displaystyle \tan^{-1}(x)+\tan^{-1}(y) = -\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\;,$when $x<0,y<0,xy>1$
