What is the sum of arcus cotangents? Say I wish to find the sum of three arcus cotangents. I wanted to start with a smaller example, say
$$\mathrm{arccot}{x} + \mathrm{arccot}{y}.$$
I know that
$$
\mathrm{arccot}{x} = 
\begin{cases} 
\arctan{\frac{1}{x}} & \text{if $x>0$} \\
\pi + \arctan{\frac{1}{x}} & \text{if $x<0$}
\end{cases}
$$
And that
$$
\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}
$$
Is this the right path to find the expression for $\mathrm{arccot}{x} + \mathrm{arccot}{y}$?
Simply consider every case of $x,y,xy$? How to get to the sum of three arcus cotangents from here?
Was also looking for a "ready" formulation online but I ultimately failed. So out of curiosity - are arcus cotangents that rarely used?


I know that  $$ \mathrm{arccot}(x) =  \begin{cases} 
\arctan{\frac{1}{x}} & \text{if $x>0$} \\ \pi + \arctan{\frac{1}{x}} &
\text{if $x<0$} \end{cases} $$

And a small addition: $$=\frac\pi2\quad\text{if }x=0.$$
Based on this definition (there are two common definitions of arccot), we have that on $\mathbb R,$ $$\arctan(x)\equiv\frac {\pi}2-\mathrm{arccot}(x).$$

and that $$\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Is this the right path to find the expression for $\mathrm{arccot }(x) + \mathrm{arccot }(y)$?

Substituting in the above identity and noting that $\arctan$ is an odd function: $$\mathrm{arccot }(x)+\mathrm{arccot }(y) = \begin{cases}\pi+\arctan\left(\dfrac{x+y}{xy-1}\right), &xy < 1 \\
\arctan\left(\dfrac{x+y}{xy-1}\right), &x>0,\; y>0,\; xy>1 \\
2\pi + \arctan\left(\dfrac{x+y}{xy-1}\right), &x<0,\; y<0,\; xy > 1.\end{cases}$$
