there exist some real $a >0$ such that $\tan{a} = a$ How can i prove that there exist some real $a >0$ such that $\tan{a} = a$ ? 
I tried compute $$\lim_{x\to\frac{\pi}{2}^{+}}\tan x=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\cos x}$$ 
We have the situation " $\frac{1}{0}$ " which leads us " $\infty$ " 
$$\lim_{x\to\frac{\pi}{2}^{-}}\tan x=\lim_{x\to\frac{\pi}{2}^{-}}\frac{\sin x}{\cos x}$$ 
We have the situation " - $\frac{1}{0}$" which tells us " $- \infty$" 
This means for any real number $y$ there exists $x_1$ and $x_2$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that 
$f(x_1)<y<f(x_2)$. 
Remember the Intermediate Value Theorem. If f is a continuous function and $f(a)<0<f(b)$ then there exist $x \in (a,b)$ such that $f(x) = 0$ 
So $f(x_1)<y<f(x_2)$ is equivalent to 
f(x1) - y < 0 < f(x2) - y (I just subtracted y from each part) 
Now we can use the Intermediate Value Theorem (applied to $(f - y)$ to say there exists an $x \in (x_1 , x_2)$ such that $f(x) - y = 0 $
or $f(x) = y$ or $\tan x = y$ 
We know $x_1 < x < x_2$ and $-\frac{\pi}{2}<x_1<x_2<\frac{\pi}{2}$ 
So we know that $-\frac{\pi}{2}<x<\frac{\pi}{2}$, then $f(x) - y = 0$.
But how can i prove it for $a >0$ such that $\tan{a} = a$ ?

Hint
You are looking for the intersection of two functions $y_1=\tan(x)$ and $y_2=x$. You also know that $\tan(x)$ has discontinuities at $x=(2k+1) \frac {\pi}{2}$ (to be more precise, as gniourf_gniourf commented, the $\tan$ is not defined at these points). So, you have an infinite number of solutions for $\tan(x)=x$ (in particular because $\tan(k\pi)=0$) and the solutions are closer and closer to the vertical asymptotes.
But, as you know, in the range $(0,\frac{\pi}{2})$ $y_2$ is the tangent to $y_1$ and there is no solution in this interval. The first solution will happen just below $\frac{3\pi}{2}$, the second still closer to $\frac{5\pi}{2}$ and so on. You also can notice that, if $x_n$ is a solution $-x_n$ is another.
Moreover, Taylor series built at $x=(2k+1) \frac {\pi}{2}$ shows then $$\tan(x) \simeq -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ so you can approximate the solutions solving for $x$  $$x= -\frac{1}{x-(2k+1)\frac{\pi}{2}}$$ which leads to $$x_k \simeq \frac{1}{4}\Big((2k+1)\pi+\sqrt {(2k+1)^2 \pi^2-16}\Big)$$ from which you could show that, for large values of $k$ $$x_k\simeq(2k+1) \frac {\pi}{2}-\frac{2}{\pi  (2 k+1)}$$
