$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ 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}

Write $a:=\arctan x$ etc. so $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan a+\tan b + (1-\tan a\tan b)\tan c}{1-\tan a\tan b - (\tan a+\tan b)\tan c}.$$If $xy\ne 1$, we can cancel $1-\tan a\tan b$ to get $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\frac{\tan (a+b)+\tan c}{1-\tan (a+b)\tan c}.$$If $\tan (a+b)\tan c\ne 1$ i.e. $(x+y)z\ne 1-xy$ i.e. $xy+yz+zx=1$, we have $$\frac{x+y+z-xyz}{1-xy-yz-zx}=\tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
