let $f : (-\frac{\pi}{2},\frac{\pi}{2}) \to \mathbb{R}$ be a function so $f(x)=\tan(x)$. I am trying to prove that $f$ is bijective. Firstly I want to show that $f$ is injection. Consider $x,y\in (-\frac{\pi}{2},\frac{\pi}{2})$ so $\tan(x)=\tan(y)$. we need to prove that $x=y$ but how?

If I want to prove that $f$ is surjective, then I want to show that for all $y\in \mathbb{R}$ there is $x\in (\frac{-\pi}{2},\frac{\pi}{2})$ so $f(x)=y$ but here I also hove some trouble to prove.

Any hints?


5 Answers 5



Suppose $\tan(x)=\tan(y)$, so $$\frac{\sin(x)}{\cos(x)} = \frac{\sin(y)}{\cos(y)}$$ Hence, $$\sin(x)\cos(y)=\sin(y)\cos(x)$$ and $$\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)=0$$ Since $x,y\in (-\pi/2,\pi/2)$, we know that $x-y \in (-\pi,\pi)$. But $\sin(z)=0$ only happens at $z=2\pi n$, so we must have $x-y=0$, i.e. $x=y$.

The other answers give the overall idea for surjectivity. Because $\cos(\pi/2)=0$ and $\sin(\pi/2)=1$, we know that $\lim_{x \to \pi/2^-}{\tan(x)}=\infty$. Likewise, $\lim_{x\to -\pi/2^+}{\tan(x)}=-\infty$. Since $\tan$ is continuous, the Intermediate Value Theorem applies to show that the image of $\tan$ must be all of $(-\infty,\infty)$.

Here's another way of getting the surjectivity, given you know that $(\cos(t),\sin(t))$ parametrizes the circle. The point $$\left( \sqrt{\frac{1}{1+r^2}},r\sqrt{\frac{1}{1+r^2}}\right)$$ is on the unit circle for any $r$, hence there is a $t$ such that $(\cos(t),\sin(t))$ gives this point. But then $$\tan t = r$$ So $\tan$ has image all of $\mathbb{R}$.

  • 1
    $\begingroup$ Just one note, you have a little mistake in the limits, $\lim_{x\to\pi/2^-}\tan(x)=\infty$ and $\lim_{x\to-\pi/2^+}\tan(x)=-\infty$ while $\lim_{x\to\pi/2^+}\tan(x)$ and $\lim_{x\to-\pi/2^-}\tan(x)$ are outside the domain $\endgroup$
    – ℋolo
    Dec 19, 2017 at 14:16
  • $\begingroup$ Oops, thanks for pointing that out! $\endgroup$
    – Hayden
    Dec 19, 2017 at 14:17
  • $\begingroup$ From the equation: $\tan(x)=\tan(y)$, can I write $x = y$ by applying $\arctan$ on both sides? $\endgroup$
    – rainman
    Dec 8, 2019 at 9:10

Show that:

  • the function is monotonically increasing on $(-\pi,\pi)$.
  • the function is not bounded above and not bounded below.
  • the function is continuous.

The first bullet ensures that the function is injective.

The second and third bullet together with intermediate value theorem ensure that the function is surjective.



Prove that $f$ is an increasing function, and that its limits at either bounds are $-\infty$ and $+\infty$, then apply the Intermediate Value theorem.


This follows from $\tan'(x)=1+\tan^2(x)$ and the fact that $\lim_{x\to\pm\pi/2}\tan x=\pm\infty$. Apply the Intermediate Value Theorem.


The cheap way would to say, that $f^{-1}\colon \mathbb{R}\to (-\frac{\pi}{2},\frac{\pi}{2})$ with $x\mapsto \arctan(x)$ is the inverse function. But I doubt, that you can use that, since it it kinda circular.

To show, that $f$ is injective, it is enough to show, that $f'(x)>0$ for every $x\in (-\frac{\pi}{2},\frac{\pi}{2})$.

It is $\tan(x)=\frac{\sin(x)}{\cos(x)}$.

Or you can do it the "common" way and use the addition theorems for $\frac{\sin(x)}{\cos(x)}=\frac{\sin(y)}{\cos(y)}$

For the surjecitivity you might apply the intermediate value theorem, after checking what happens for the limits.


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