Proof regarding $\operatorname{arccot}$, where am I mistaken?

Question

We know that if $\cot(-\theta)=-\cot(θ)=\cot(\pi-θ)=x$, then

1. $ \cot(-θ) = x \quad\implies\quad -θ = \newcommand{\arccot}{\operatorname{arccot}}\arccot(x) \quad\implies\quad θ = -\arccot(x)$

2. $-\cot(θ) = x \quad\implies\quad \cot(θ) = -x \quad\implies\quad θ = \arccot(-x)$

3. $\cot(π-θ) = x \quad\implies\quad π-θ = \arccot(x) \quad\implies\quad θ = π-\arccot(x)$

Now we get that $$\arccot(-x)=π-\arccot(x)=-\arccot(x)$$ because my book has $\arccot(-x) = π-\arccot(x)$ and I added $\arccot(-x) = π-\arccot(x)=-\arccot(x)$. This leads to misleading statement $pi=0$, but my steps are not wrong (I guess). Where am I mistaken?

Answer 1

There are a few incorrect statements in your reasoning. First let's establish what has been said thoroughly in the comments. The $\cot()$ function takes any real number besides a multiple of $\pi$ as its input and can return any real number as its output. The $\newcommand{\arccot}{\operatorname{arccot}}\arccot()$ function is defined to take any real number as its input and return a number in the range $(0,\pi)$ as its output. It is necessary to restrict the range of $\arccot()$ like this because otherwise it wouldn't be a function (remember that each input needs a unique output). Because of this restriction, some sensible statements about $\arccot()$ can be made, that aren't actually true. For example, one of your statements above is

$$\cot(-\theta) = x \quad\implies\quad -\theta = \arccot(x)\;.$$

This isn't true. For example $\cot(-\frac{37}{2}\pi) = 0$, but $\arccot(0) \neq -\frac{37}{2}\pi$ because $\arccot(0)$ has to be in the range $(0,\pi)$. In fact, many values make $\cot()$ return $0$, but only one of those values can be returned by $\arccot(0)$. You also made the statement

$$\arccot(-x) = -\arccot(x)\;.$$

This isn't ever true. Because $\arccot()$ must always return a positive number we have $\arccot(\_) \neq -\arccot(\_)$ for any choice of inputs.