$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$? 
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?

This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $\tan^{-1}x+\tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z$ ?
\begin{align}
\tan^{-1}x+\tan^{-1}y+\tan^{-1}z&=
\begin{cases}\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy < 1 \\[1.5ex] 
\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y>0 \\[1.5ex] 
-\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y<0 \end{cases}\\
&=
\end{align}

We have that by addition formula
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$
then
$$(\tan^{-1}x+\tan^{-1}y)+\tan^{-1}z=\tan^{-1}\frac{\frac{x+y}{1-xy}+z}{1-\frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)


*

*$1-xy=0$
and is certainly not well defined for


*

*$1-xy-yz-zx=0$
