Why does applying $\tan$ to $\arctan x$ make it $x$?

Question

Through doing calculus problems I have found that $\tan(\arctan x)$ seems to be equal to $x$. I can't seem to understand why. Can someone please explain this to me? Also, does this rule apply to anything else? This rule, as in, something of the inverse of something of $x$ is equal to $x$.

Answer 1

Yes. $$f(f^{-1}(x))=x= f^{-1}(f(x))$$ Note that $\arctan=\tan^{-1}$.

Simpler example: Take $f(x)=x^2$. Then $x=\sqrt{f(x)}\implies f^{-1}(x)=\sqrt x$ and so $$f(f^{-1}(x))=\sqrt{x}^2=x$$

Answer 2

Just adding to what has already been said, since $\tan$ is periodic so its inverse $\arctan$ is defined only as the reciprocal of its restriction on $\left( \frac{-\pi}{2} , \frac{\pi}{2} \right)$. So: $$ \tan(\arctan(x))=\arctan(\tan(x))=x $$ is true for $x\in \left( \frac{-\pi}{2} , \frac{\pi}{2} \right)$ only as $\arctan(\tan(x))$ is not always $x$ ($\arctan$ only has values in $\left( \frac{-\pi}{2} , \frac{\pi}{2} \right)$)

Edit (all credit goes to Jyrki Lahtonen): $\tan(\arctan y)=y$ holds for all $y\in \mathbb{R}$. We view $\tan$ as a function from $\left( \frac{-\pi}{2} , \frac{\pi}{2} \right) \to \mathbb{R}$ and $\arctan$ as a function from $\mathbb{R} \to \left( \frac{-\pi}{2} , \frac{\pi}{2} \right)$. With the domains and codomains defined that way they are inverses to each other.