Proof: $\tan(x)$ is surjective from $(-\pi/2,\pi/2)$ onto $\mathbb R$ To prove $\tan(x)$ defined on $]-π/2;π/2[$ is injective I take the derivative of $\tan(x)$ to get $\sec(x)^2$.
This shows that $\tan(x)$ is monotonic (strictly) increasing which implies it is injective.
However how do I show it is surjective ? That every single real number corresponds to some number in the domain of $\tan(x)$ ?

Given that $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\rightarrow \mathbb{R}, f(x) = \tan(x)$.
Proof of Surjectivity
Suppose $\alpha \in \mathbb{R}$. We have to prove that $\exists \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$  for which $f(\beta) = \alpha$. 
We will use the Intermediate Value Theorem to prove that. 
Intermediate Value Theorem
Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous, and assume that $s$ lies between $f(a)$ and $f(b)$. Then $\exists$ at least one $c \in [a, b]$ such that $f(c) = s$.
I omit the proof of the fact that $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\rightarrow \mathbb{R}, f(x) = \tan(x)$ is continuous. 
Let $-\infty < \alpha < \infty$, where $\infty = \lim_{x \to {\frac{\pi}{2}}^-} f(x)$ and, $-\infty = \lim_{x \to {-\frac{\pi}{2}}^+} f(x)$. Note that $\lim_{x \to {\pm \frac{\pi}{2}}} f(x)$ do not exists.
Then by the Intermediate Value Theorem, $\exists$ at least one $\beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $f(\beta) = \alpha$. 
It follows that $f$ is surjective.
This answer is inspired by the comments above. 
