Why is $\operatorname{arccot} x$ not $\arctan \frac{1}{x}$ when $x<0$?

Question

$\operatorname{arccsc}(x) = \arcsin \frac{1}{x}$ , and $\operatorname{arcsec}(x) = \arccos \frac{1}{x}$.

Why doesn't $\operatorname{arccot}(\cdot)$ behave the same..?

Answer 1

It is just a matter of convention. Refer to these illustrations as you read the following:

1. Wolfpham Alpha and many (most?) texts define $\mathrm{arccot}(x)$ over the domain $\mathbb R$ by inverting the discontinuous $\left(-\frac{\pi}2,0\right)\cup\left(0,\frac{\pi}2\right]$ section of $\cot(x).$ Consequently, $$\mathrm{arccot}(x)=\begin{cases} \arctan\left(\frac1x\right) &\text{ when }x\neq0;\\ \frac {\pi}2 &\text{ when }x=0, \end{cases}$$ and has range $$\left(-\frac{\pi}2,0\right)\cup\left(0,\frac{\pi}2\right].$$
2. On the other hand, Desmos defines $\mathrm{arccot}(x)$ over the domain $\mathbb R$ by inverting the continuous $(0,\pi)$ section of $\cot(x).$ (Wikipedia says that this is the usual definition.) Consequently, $$\mathrm{arccot}(x)=\begin{cases} \arctan\left(\frac1x\right)+\pi &\text{ when }x<0;\\ \frac {\pi}2 &\text{ when }x=0;\\ \arctan\left(\frac1x\right) &\text{ when }x>0, \end{cases}$$ and has range $$\large\left(0,\pi\right).$$

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Jean-Claude points out that while the first definition above gives $$\mathrm{arccot}(x)=\begin{cases} -\frac {\pi}2-\arctan(x) &\text{ when }x<0;\\ \frac {\pi}2-\arctan(x) &\text{ when }x\geq0, \end{cases}$$ the second definition has $\mathrm{arccot}$ and $\arctan$ as reflections of each other in the line $y=\frac{\pi}4$, so^҂ gives the friendlier $$\mathrm{arccot}(x)=\frac {\pi}2-\arctan(x)\quad\text{ on }\mathbb R.$$

All things considered $\big($clean relationship with $\arctan\left(\frac1x\right),$ range centred about $0$   versus   clean relationship with $\arctan(x),$ continuous$\big),$ I prefer the second definition to the first.

^҂An analytical argument (based on the second definition) is this:
$\quad\arctan(x)$ and $\mathrm{arccot}(x)$ are both continuous on $\mathbb R,$ so $$\arctan(x)+\mathrm{arccot}(x)$$—which has derivative $0$ on $\mathbb R$ and value $(\frac{\pi}4+\frac{\pi}4=\frac{\pi}2)$ at $(x=1)$—is also continuous on $\mathbb R.$ As such, this sum identically equals $\frac{\pi}2.$

Answer 2

To elaborate on Alan's comment:

Since the functions $\tan$ and $\cot$ have period $\pi$, given a real number $x\in\mathbb R$, there is no "correct" way to choose its inverse $\arctan(x)$ or $\operatorname{arccot}(x)$. For instance, $\arctan(0)$ could be $0$, or $\pi$, or any other multiple of $\pi$.

However, let us now choose an (arbitrary) value for $\arctan(0)$, say $0$. Then, there is only one way to define $\arctan$ while having it be continuous. This choice makes the range of $\arctan$ be $(-\pi/2,\pi/2)$, which is exactly the interval Alan mentioned.

Similarly, we usually choose $\operatorname{arccot}(0)=\pi/2$ and let it be continuous. This defines $\operatorname{arccot}$ uniquely, and makes its range $(0,\pi)$.

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The above explains how the definitions of $\arctan(x)$ and $\operatorname{arccot}(x)$ came to be. Now, why does this weird discontinuity exist?

This is because we are considering $\arctan(1/x)$, and $1/x$ is discontinuous at $x=0$. Thus, it makes sense that there is a discontinuous "jump" at $x=0$.