Why $x>0$ for $\tan^{-1}\big(\tfrac{1}{x}\big)=\cot^{-1}x$

Question

$$ \tan^{-1}\big(\tfrac{1}{x}\big)=\cot^{-1}x, \quad x>0 $$

I understand the simple proof $$ y=\cot^{-1}x\implies \cot y=x\implies\tfrac{1}{x}=\tan y\\ \tan^{-1}\big(\tfrac{1}{x}\big)=\tan^{-1}\big(\tan y\big)=y=\cot^{-1}x $$ From the domains of $\tan^{-1}$ and $\cot^{-1}$, $$ \tfrac{1}{x}\in\mathbb{R} \quad\&\quad x\in\mathbb{R}\\\implies {x}\in\mathbb{R}-\{0\} \quad\&\quad x\in\mathbb{R}\implies x\in\mathbb{R}-\{0\} $$ I can understand $x\neq{0}$, but how come the condition $x>0$ ?

My understanding

For $\sin^{-1}$,

$$ \sin^{-1}\big(\tfrac{1}{x}\big)=\csc^{-1}x,\quad x\leq{-1}\text{ or }x\geq{1} $$

$$ -1\leq\tfrac{1}{x}\leq1 \quad\&\quad x\leq-1\text{ or }x\geq1\\-1\leq\tfrac{1}{x}\leq1 \implies -1\leq\tfrac{1}{x}<0\text{ or }0\leq \tfrac{1}{x}\leq1\implies x\leq{-1}\text{ or }x\geq 1\\ x\leq{-1}\text{ or }x\geq 1\quad\&\quad x\leq{-1}\text{ or }x\geq 1\implies x\leq{-1}\text{ or }x\geq 1 $$ from the domains of the functions $\sin^{-1}$ and $\csc^{-1}$.

Note: I am only considering the pricipal value branch.ie., $$ \tan^{-1}:\mathbb{R}\to\Big(-\pi/2,\pi/2\Big)\\ \cot^{-1}:\mathbb{R}\to\Big(-\pi,\pi\Big) $$

Answer 1

Usually the branches of $\cot^{-1}$ are chosen to be continous at zero, see the Wiki picture. You have $\cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x).$ This is sometimes called the continuous inverse circular cotangent.

There is another choice called the sign symmetric inverse circular cotangent with $\cot^{-1}(x) = \tan^{-1}(\frac 1x), $ and here your domain condition $x\ne 0$ applies.

Answer 2

We have the well-known relation: $$\arctan x+\arctan\frac1x=\begin{cases}\phantom{-}\dfrac\pi2&\text{ if }x>0,\\[1ex]-\dfrac\pi2&\text{ if }x<0,\end{cases}$$ so, if $x>0$, $\;\arctan\dfrac1x=\dfrac\pi2-\arctan x$, so that $$\cot\Bigl(\arctan\frac1x\Bigr)=\cot\Bigl(\frac\pi2-\arctan x\Bigr)=\tan(\arctan x)=x.$$

Answer 3

For $x>0$ $$cot\tan^{-1}\big(\tfrac{1}{x}\big)=\frac{1}{tan\tan^{-1}\big(\tfrac{1}{x}\big)}=x$$