Range of $\operatorname{arccot}(x)$

Question

Why don't we choose range of $\operatorname{arccot}(x)$ to be $\left(\frac{-\pi}{2},\frac{\pi}{2}\right]-\{0\}$ instead of $(0,\pi)$ so that it becomes little easier to work with arctan and arccot if they have similar range.

Answer 1

You can do that. Some people do, including Mathematica, MATLAB, and MuPAD. As the latter implementation notes:

The inverse cotangent function is multivalued. The MuPAD arccot function returns the value on the main branch. The branch cut is the interval $[- i, i]$ on the imaginary axis. Thus, arccot returns values, such that $y = \mathrm{arccot}(x)$ satisfies $-\frac\pi2<\Re(y)\leq\frac\pi2$ for any finite complex $x$.

... The values jump when the arguments cross a branch cut.

... Note: MuPAD defines arccot as $\mathrm{arccot}(x) = \mathrm{arctan}(1/x)$ ... As a consequence of this definition, the real line crosses the branch cut, and arccot has a jump discontinuity at the origin.

So your point about the ranges of arctan and arccot was important to that author. But they also felt it was important to warn the reader about the discontinuity.

There's a more complete overview of implementations here: http://www.intmath.com/blog/mathematics/which-is-the-correct-graph-of-arccot-x-6009