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

Let

• $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$


• $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$

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
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$

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
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
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

$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$

Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ 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).

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

• $1-xy-yz-zx=0$

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 $2pi$) 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 $tanphi=b/a$.

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 $3pi/4$ 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.