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

Question

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

Answer 1

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

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

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

Answer 2

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.

Answer 3

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

• $1-xy-yz-zx=0$

Answer 4

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).