Prove $\tan$ function is a bijection 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?
 A: 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.
A: Hint:
Prove that $f$ is an increasing function, and that its limits at either bounds are $-\infty$ and $+\infty$, then apply the Intermediate Value theorem.
A: 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.
A: Injectivity:
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 $\mathbb{R}$.
A: 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.
