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 $\tan(x)$ ?
 A: 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
$$
\lim_{\theta\to\pi/2^-}\tan(\theta)=\infty\qquad\text{and}\qquad\lim_{-\theta\to-\pi/2^+}\tan(-\theta)=-\infty.\qquad\approx\blacksquare
$$
A: The above use of the intermediate value theorem uses $\overline{\mathbb R}$ to accomplish its task, but also seems to beg the question by asserting the one-sided limits of tangent at $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.  Here is a proof that does not use either geometry or $\overline{\mathbb R}$ The proof will however use various derivative rules including L'Hospital and Intermediate Value Theorem (all of which can clearly be derived in a noncircular manner)
$\forall x\in\mathbb R$, it is clear there $\exists k\in\mathbb N$ at which $2^k \ge x$ (proof found elsewhere).  We can use this fact to show tangent is surjective on $\mathbb R$ on the $\mathcal D_{\tan}(-\frac{pi}{2},\frac{\pi}{2})$ without resorting to geometry.
The derivative of $f(x) = \frac{tan(\frac{\pi/2+x}{2})}{tan(x)}$ is, after plenty of simplification, $\frac{1}{cos(\frac{\pi/2+x}{2})sin(x)} (\frac{cos(x)}{2cos(\frac{\pi/2+x}{2})} - \frac{sin(\frac{\pi/2+x}{2})}{sin(x)})$.  This function can be seen to be negative on the domain $(\pi/4,\pi/2)$.
The limit of $f$ as $x$ approaches $\pi/2$ is 2, and is easily findable using the chain rule as well as L'Hopital rule.  Then it is clear since the derivative is negative and the minimum is 2, that all values of $x$ on the open interval $(\pi/4,\pi/2)$ have $f(x) \ge 2$.
This is when the pieces start to fall into place.  Let's form the sequence $a_i$ defined by $a_1 = \pi/4$, $a_j = (\pi/2 + a_{j-1})/2$ for all other $j$.  We've already shown that for all $j > 1$, $\frac{\tan(a_j)}{\tan(a_{j-1})} \ge 2$.  It is therefore clear that for all real $r>=1$, there is some $i_1$ where $tan(a_{i_1}) \ge 2^{{i_1}-1} \ge r$ (from the above mentioned theorem).  
Since $\tan(-x) = -\tan(x)$, the same principle can be applied to any negative real $r <= -1$ to find an $i_2$ where $tan(-a_{i_2}) \le -2^{{i_2}-1} \le r$. These two values are appropriate for use in the intermediate value theorem to show there must exist some value $k$ in $(a_{i_2},a_{i_1})$ where $\tan(k) = r$.
A: 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. 
