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 $\tan{a} = a$ ?

Use the intermediate value theorem on $\tan x -x$ after plugging in $x={3\pi\over 4}$ and $x={35\pi\over 24}$.
I chose those specific values because (in theory) you can compute them by hand using half-angle formulae and it's more constructive, but one can use the definition of unboundedness near $3\pi/2$ and the boundedness of the identity function on any compact set to make things slicker.
