Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

I know that $\displaystyle \cot(x)=\frac{1}{\tan(x)}$ and $\space \displaystyle \arctan(x)=\tan(x)^{-1}=\frac{1}{\tan(x)}$

What is the difference between these two function?

Is $\cot(x)$ the reciprocal function of $\space \tan(x) \space$ and $\arctan(x)$ is the inverse function of $\tan(x)$?

And, so the assumption that $\space \displaystyle \arctan(x)=\tan(x)^{-1}=\frac{1}{\tan(x)}$, is incorrect?

share|improve this question

marked as duplicate by Guess who it is., user1729, Amzoti, Stefan Hansen, Lord_Farin Jun 11 '13 at 12:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$f^{-1}(x)\neq f(x)^{-1}$. It is in some ways unfortunate notation. –  Jonas Meyer Jun 11 '13 at 11:32
I suppose your confussion stems from the fact that some pretty well-known scientific calculators (e.g., at least some of the Casio ones) denote by $\,\sin^{-1}\,,\,\cos^{-1}\,,\,\tan^{-1}\,$ the inverse trigonometric functions...a very poor, sloppy choice by those Casio guys, and since these calculators are very widely used by high school students and these don't usually have much idea what's going on, the confussion persists later... –  DonAntonio Jun 11 '13 at 11:50

3 Answers 3

up vote 2 down vote accepted

There is a mistake. And the main problem is the notation. If we have a function $f$ that is 1-1 then we can think of it's inverse function $f^{-1}$. This notation can mislead people to the wrong impression that $f^{-1}(x)=1/f(x)$. That's not true! The function $f^{-1}$ is the function such that $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$. For instance, let $f : \Bbb R \to \Bbb R$ be given by: $f(x)=\lambda x$. In that case, $f$ is obviously 1-1 with inverse $f^{-1}(x)=x/\lambda$. Notice that $f^{-1}(x) \neq 1/f(x)$.

In that case we define two things for $\tan$: the reciprocal function $\cot $ that is really defined by $\cot(x) = 1/\tan (x)$ and the inverse function $\arctan$ given by the property I've mentioned above. Take a look on my answer here about the same doubt involving $\sec$, it's the same issue and it may help you.

Good luck.

share|improve this answer

Yes. $\cot x$ is the reciprocal, $\arctan x$ is the (principal) inverse, and $\arctan x=\frac1{\tan x}$ is incorrect.

share|improve this answer

Think about this;

$tan(0)=0$ and so one would expect that the inverse of this function 'arctan' to satisfy $arctan(0)=0$. However, $cot(0) = \frac{1}{tan(0)}$ which of course is a 'division by zero' error.

It's always a good idea if you are confused with the notation/trig functions is to just try a few notable points. e.g. 0, $\frac{\pi}{2}$, $\pi$ etc...

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.