Range of $\cot^{-1}(x)$ S. L. Loney mentions the range of $\cot^{-1}(x)$ to be $[-\pi/2,0)\cup(0,\pi/2]$ whereas my textbook (my coaching institute package) mentions it to be $(0,\pi)$. May I know which one is the correct convention? I have referred many resources which show the range to be  $(0,\pi)$. Are there any advantages of defining it this way over the previous one?
 A: 
which one is the correct convention?

As already pointed out, as for any other inverse function defined over a restriction, no one is the correct convention and either is acceptable.
The two ranges usually adopted for $\cot^{-1}(x)$ are the more convenient or more simple or more natural.
Indeed we could also define the range in infinitely many other ways for $c\in(0,\pi/2]$
$$  [-c,0)\cup(0,\pi-c]$$
and each of them would be a proper choice to define $\cot^{-1}(x)$.

Are there any advantages of defining it this way over the previous one?

Each choice may have advantages depending on the context, notably:

*

*By the choice $[-\pi/2,0)\cup(0,\pi/2]$ we have that $\cot^{-1}(x)$ is an odd function discontinuous at $x=0$ since

$$\lim_{x\to 0^-} \cot^{-1}(x)=-\frac \pi 2 \neq \lim_{x\to 0^+} \cot^{-1}(x)=\frac \pi 2$$
This is for example the function we find on Wolfram


*

*By the choice $(0,\pi)$ we have that $\cot^{-1}(x)$ is no more an odd function but it is continuous.

Moreover in this case the following holds $\forall x \in \mathbb R$
$$\cot^{-1}(x)=\frac \pi 2 - \tan^{-1}(x)$$

(image credits)
A: It's better to take the range as $(0,\pi)$ as it gives continuous graph.
