Trying to find $\cot^{-1}x-\cot^{-1}y$ for $\forall$ $x,y$

Question

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 $\cot^{-1}x+\cot^{-1}(y)$.

Answer 1

$$0<\cot^{-1}x+\cot^{-1}y<2\pi$$

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 $x \in [-1,1]$?