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

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.,
$$
\tan^{-1}:\mathbb{R}\to\Big(-\pi/2,\pi/2\Big)\\
\cot^{-1}:\mathbb{R}\to\Big(-\pi,\pi\Big)
$$
 A: 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.
A: 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
$$\cot\Bigl(\arctan\frac1x\Bigr)=\cot\Bigl(\frac\pi2-\arctan x\Bigr)=\tan(\arctan x)=x.$$
A: For $x>0$
$$cot\tan^{-1}\big(\tfrac{1}{x}\big)=\frac{1}{tan\tan^{-1}\big(\tfrac{1}{x}\big)}=x$$
A: $$
\tan^{-1}:\mathbb{R}\to{\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)}\quad\&\quad\cot^{-1}:\mathbb{R}\to{\big(0,\pi\big)}
$$
Taking,
$$
\cot^{-1}x=\alpha\implies\cot\alpha=x, \text{ where }0<\alpha<{\pi}\\\implies\tan\alpha=\frac{1}{x}=\tan\Big[\tan^{-1}\frac{1}{x}\Big]\\\implies\boxed{\tan^{-1}\frac{1}{x}=n\pi+\alpha}
$$
If $0<\alpha<\tfrac{\pi}{2}$,
$$
\tan^{-1}\frac{1}{x}=\alpha\implies \tan^{-1}\frac{1}{x}=\cot^{-1}x
$$
$0<\alpha<\tfrac{\pi}{2}\implies \infty>\cot\alpha>0\implies\infty>x>0\implies\color{red}{x>0}$

$$
\tan^{-1}\frac{1}{x}=\cot^{-1}x, \quad\text{ if }x>0
$$

If $\tfrac{\pi}{2}<\alpha<\pi$,
$$
\tan^{-1}\frac{1}{x}=-\pi+\alpha\implies \tan^{-1}\frac{1}{x}=-\pi+\cot^{-1}x
$$
$\tfrac{\pi}{2}<\alpha<\pi\implies 0>\cot\alpha>-\infty\implies 0>x>-\infty\implies\color{red}{x<0}$

$$
\tan^{-1}\frac{1}{x}=-\pi+\cot^{-1}x, \quad\text{ if }x<0
$$

