$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ 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}
Best Answer
Let
By addition formula of tangent function, we have $$\tan L(x,y,z) = \tan R(x,y,z)$$ Since $\tan \theta$ is a periodic function with period $\pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)\pi$$
Since $\tan^{-1}\theta$ maps $\mathbb{R}$ into $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have $$|L(x,y,z)| < \frac{3\pi}{2} \land |R(x,y,z)| < \frac{\pi}{2}\quad\implies\quad N(x,y,z) \in \{ 0, \pm 1 \}$$
Since $\tan^{-1} \theta$ is a continuous function for all $\theta$, $N(x,y,z)$ will be constant over those domain where $xy+yz+zx \ne 1$. Notice $$xy+yz+zx = 1 \iff 3\left(\frac{x+y+z}{\sqrt{3}}\right)^2 - ( x^2 + y^2 + z^2 ) = 2$$ is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + \begin{cases} \pi, & 1 < xy+yz+zx \land x+y+z > 0\\ 0, & 1 > xy+yz+zx\\ -\pi & 1 < xy+yz+zx \land x+y+z < 0 \end{cases} $$