<s>[INST] Trying to find $\cot^{-1}x-\cot^{-1}y$ for $\forall$ $x,y$ Suppose we want to calculate $\cot^{-1}x-\cot^{-1}y$ for $\forall~x,y$
$$\cot^{-1}x-\cot^{-1}y=\theta\tag{1}$$
Let's find range of $\theta$, assuming $x$ and $y$ to be independent variables
$$\theta\in(-\pi,\pi)$$ 
Taking $\cot$ on both sides of equation $1$
$$\dfrac{\cot(\cot^{-1}x)\cot(\cot^{-1}y)+1}{\cot(\cot^{-1}y)-\cot(\cot^{-1}x)}=\cot\theta$$
$$\dfrac{xy+1}{y-x}=\cot\theta$$
Taking $\cot^{-1}$ on both sides
$$\cot^{-1}\dfrac{xy+1}{y-x}=\cot^{-1}(\cot\theta)$$
$$\cot^{-1}(\cot\theta)=\begin{cases}
    \pi+\theta,&-\pi<\theta<0 \\
    \theta,&0<\theta<\pi \\
  \end{cases}$$
So
$$\theta=\begin{cases}
    -\pi+\cot^{-1}\dfrac{xy+1}{y-x},&-\pi<\theta<0 \\
    \cot^{-1}\dfrac{xy+1}{y-x},&0<\theta<\pi \\
  \end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
    -\pi+\cot^{-1}\dfrac{xy+1}{y-x},&-\pi<\cot^{-1}x-\cot^{-1}y<0 \\
    \cot^{-1}\dfrac{xy+1}{y-x},&0<\cot^{-1}x-\cot^{-1}y<\pi\\
  \end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
    -\pi+\cot^{-1}\dfrac{xy+1}{y-x},&\cot^{-1}x-\cot^{-1}y\in(-\pi,0) \\
    \cot^{-1}\dfrac{xy+1}{y-x},&\cot^{-1}x-\cot^{-1}y\in(0,\pi)\\
  \end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
    -\pi+\cot^{-1}\dfrac{xy+1}{y-x},&\dfrac{xy+1}{y-x}\in(-\infty,\infty) \\
    \cot^{-1}\dfrac{xy+1}{y-x},&\dfrac{xy+1}{y-x}\in(-\infty,\infty)\\
  \end{cases}$$
So finally we can write 
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
    -\pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
    \cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
  \end{cases}$$
We can also derive for $\cot^{-1}x+\cot^{-1}y$ in the following way
Adding $\pi$ to both sides
$$\cot^{-1}x+\pi-\cot^{-1}y=\begin{cases}
    \cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
    \pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
  \end{cases}$$
$$\cot^{-1}x+\cot^{-1}(-y)=\begin{cases}
    \cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
    \pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
  \end{cases}$$
Replacing $y$ with $-y$
$$\cot^{-1}x+\cot^{-1}(y)=\begin{cases}
    \cot^{-1}\dfrac{-xy+1}{-y-x}, &x>-y\\
    \pi+\cot^{-1}\dfrac{-xy+1}{-y-x}, &x<-y\\
  \end{cases}$$
$$\cot^{-1}x+\cot^{-1}(y)=\begin{cases}
    \cot^{-1}\dfrac{xy-1}{y+x}, &x+y>0\\
    \pi+\cot^{-1}\dfrac{xy-1}{y+x}, &x+y<0\\
  \end{cases}$$
Is it correct, I am asking because I am not sure about it because in text-books I just find $\cot^{-1}x-\cot^{-1}y=\cot^{-1}\dfrac{xy+1}{y-x}$ and I was not finding formula for $\cot^{-1}x+\cot^{-1}(y)$.
 [/INST] $$0<\cot^{-1}x+\cot^{-1}y<2\pi$$
So, $\cot^{-1}x+\cot^{-1}y$ will be $\cot^{-1}\dfrac{xy-1}{x+y}$  if $\cot^{-1}x+\cot^{-1}y<\pi$
$\iff\tan^{-1}x+\tan^{-1}y>0\iff\tan^{-1}x>-\tan^{-1}y=\tan^{-1}(-y)$
As $\tan^{-1}x$ is strictly increasing in $\left(-\dfrac\pi2,\dfrac\pi2\right),$ we need $x>-y\iff x+y>0$
So, $\cot^{-1}x+\cot^{-1}y$ will be $\pi+\cot^{-1}\dfrac{xy-1}{x+y}$  if $\pi<\cot^{-1}x+\cot^{-1}y<2\pi$
$\iff-\pi<\tan^{-1}x+\tan^{-1}y<0\iff x+y<0$
Now replace $y$ with $-z$ 
and use  $\cot^{-1}(-z)=\pi-\cot^{-1}z$ like How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$? 
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