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

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., $$ \tan^{-1}:\mathbb{R}\to\Big(-\pi/2,\pi/2\Big)\\ \cot^{-1}:\mathbb{R}\to\Big(-\pi,\pi\Big) $$

Sooraj S
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  • What is the context where you found the condition listed? – abiessu Jan 12 '18 at 14:33
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    Probably the branch of $\arctan$ used has values in $(-\pi/2,\pi/2)$ and the used branch of $\operatorname{arccot}$ has values in $(0,\pi)$. – Daniel Fischer Jan 12 '18 at 14:35
  • See https://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function – lab bhattacharjee Jan 12 '18 at 14:36
  • @DanielFischer What if we have used principal branch for both $\tan^{-1}$ and $\cot^{-1}$ ? – Sooraj S Jan 12 '18 at 14:58
  • What is the principal branch? See my answer. – gammatester Jan 12 '18 at 15:01
  • @ss1729 Those are what are usually called the principal branches. – Daniel Fischer Jan 12 '18 at 15:03
  • @DanielFischer i'm srry didnt get ur point. i mean the ranges $\tan^{-1}$ is taken as $\big(-\pi/2,\pi/2\big)$ and for $\cot^{-1}$ it is $\big(0,\pi\big)$. Thats wht principal branches means right?.. how this affect the $x>0$ condistion here? – Sooraj S Jan 12 '18 at 15:58
  • To have the relation $\operatorname{arccot} x = \arctan \frac{1}{x}$, the value of both sides must lie in the intersection of the ranges of the two functions, that is $(0,\pi/2)$. But $\arctan \varphi \in (0,\pi/2) \iff \varphi > 0$ for the principal branch. – Daniel Fischer Jan 12 '18 at 16:02
  • @DanielFischer thnx. So can i say this, $0<y=\tan^{-1}\big(\tfrac{1}{x}\big)=\cot^{-1}(x)<\tfrac{\pi}{2} \implies 0<\tan{y}=\tfrac{1}{x}<\infty\implies x>0$. – Sooraj S Jan 13 '18 at 09:57
  • It has somehow been expressed in other comments and answers but I like to repeat that the relation $\tan^{-1}\big(\tan y\big)=y$ is FALSE in general (for example, because the LHS is $\pi$-periodic while the RHS is not). The relation is true only if one assume that $y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. – Taladris Jan 15 '18 at 10:32
  • @Taladris I am only considering the pricipal value branch. srry i should have mentioned in the question itself, rather than in the comments. – Sooraj S Jan 15 '18 at 10:38

4 Answers4

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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.

gammatester
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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 $$\cot\Bigl(\arctan\frac1x\Bigr)=\cot\Bigl(\frac\pi2-\arctan x\Bigr)=\tan(\arctan x)=x.$$

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

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$$ \tan^{-1}:\mathbb{R}\to{\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)}\quad\&\quad\cot^{-1}:\mathbb{R}\to{\big(0,\pi\big)} $$ 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 $\tfrac{\pi}{2}<\alpha<\pi$, $$ \tan^{-1}\frac{1}{x}=-\pi+\alpha\implies \tan^{-1}\frac{1}{x}=-\pi+\cot^{-1}x $$ $\tfrac{\pi}{2}<\alpha<\pi\implies 0>\cot\alpha>-\infty\implies 0>x>-\infty\implies\color{red}{x<0}$

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

Sooraj S
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