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


This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors. 

Let $\alpha=\arctan x$, $\beta=\arctan y$ and $\gamma=\arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $\alpha+\beta+\gamma$ is the argument 
(up to an integer multiple of $2\pi$) of the product 
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $\phi$ of a complex number $a+ib$ satisfies $\tan\phi=b/a$. 
The claim follows from this. 

Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $\pi$. For example, if $x=y=z=1$ we have $\arctan x=\arctan y=\arctan z=\pi/4$ giving $3\pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $\arctan(-1)=-\pi/4$ on the right hand side.

