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