Proof of $\operatorname{arccot}(x)= \arctan(\frac{1}{x})$

Question

$\newcommand{\ac}{\operatorname{arccot}}$ Here is my proof:

$ \theta = \ac(x)$

$ \cot \theta = x $

$ \displaystyle \tan \theta = \frac{1}{\cot \theta} = \frac{1}{x}$

$ \displaystyle \theta = \arctan \frac{1}{x}$

But based on the graph, $\ac(x) = \arctan(\frac{1}{x})$ is only satisfied when $ x > 0$. What is wrong with my proof? What is the error occurred inside a proof?

Answer 1

There are two conventions for principal value:

1. $\operatorname{arccot} x \in (-\pi/2, \pi/2]$ with a discontinuity at zero.
2. $\operatorname{arccot} x \in (0, \pi)$

See https://mathworld.wolfram.com/InverseCotangent.html

It turns out that $\operatorname{arccot} x = \arctan(1/x)$ for all $x\ne 0$ for one convention of principal value.

If you want to adapt or enforce a particular range for $\operatorname{arccot} x$, then $\operatorname{arccot} x = \arctan(1/x) + k \pi$ for some integer $k$.

Here is what Mathematica thinks:

 Plot[{ArcTan[1/x] + .01, ArcCot[x]}, {x, -7 Pi, 0}, Frame -> True, 
  PlotRange -> All]

You can rotate this second plot $90^\circ$ and define your range anyway you want up to a multiple of $\pi.$

Answer 2

HINT.-Using a figure of a right triangle, it is evident that for $0\lt \theta\lt \dfrac{\pi}{2}$ one has the proposed equality.

from which the equality is true for all $x\gt 0$.

However, for $x\lt0$ the equality is not true because, for example,

$$\operatorname{arccot}(-2)\approx2.6779\\\arctan(-\dfrac 12)\approx-1.1071$$

Answer 3

Long comment:

The first problem with $$\arctan\left(\frac1x\right)=\operatorname{arccot}(x)$$ is that $\arctan$ is an odd function, while $\operatorname{arccot}$ isn't.

Indeed, $\forall x\in\Bbb R$, the following holds:

$$\arctan(x)+\operatorname{arccot}(x)=\frac\pi2\implies\operatorname{arccot}(x)=\frac\pi2-\arctan(x)$$ Hence, if $x>0$, then $$\arctan\left(\frac1{-x}\right)=-\arctan\left(\frac1x\right)=-\operatorname{arccot}(x)=-\operatorname{arccot}(-(-x))$$ and we cannot just move the sign under the argument. Therefore, $$\arctan\left(\frac1x\right)=\begin{cases}-\operatorname{arccot}(-x), &x<0\\\operatorname{arccot}(x),&x>0\end{cases}$$