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 $\tan(x)$ ?
I can see that a rigorous analytical proof for the surjectivity of $f\colon(-\pi/2,\pi/2)\to\mathbb{R}$, where $f(x)=\tan(x)$, is a way far off. However, I wanted to give a picture of what @achillehui mentioned in a comment, as it is rather beautiful in my opinion:

$\color{white}{Put it in the center!!!!!!!}$

Given the picture above, we note that
$$
\tan(\theta)=\frac{x}{1}=x\qquad\text{and}\qquad\tan(-\theta)=-\frac{x}{1}=-x.
$$
Hence, we can "clearly" see that
$$
\lim_{\theta\to\pi/2^-}\tan(\theta)=\infty\qquad\text{and}\qquad\lim_{-\theta\to-\pi/2^+}\tan(-\theta)=-\infty.\qquad\approx\blacksquare
$$
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 $-\infty &lt; \alpha &lt; \infty$, 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.