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**Defunkt** Don't really think there is any need for Lagrange here...

Let $\displaystyle f(x) = tan(x) -x, ~ x \in (0, \frac{\pi}{2})$, then $\displaystyle f'(x) = \frac{1}{cos^2(x)} - 1$

Now, note that:

1. $\displaystyle f(0) = tan(0) - 0 = 0$

2. Since for any $\displaystyle x \in (0, \frac{\pi}{2}), ~ 0 < cos(x) < 1 \Rightarrow 0 < cos^2(x) < 1 \Rightarrow \frac{1}{cos^2(x)} > 1$, we get that $\displaystyle f'(x) > 0 ~\forall x \in (0, \frac{\pi}{2})$.

These two facts, along with the continuity of $\displaystyle f(x)$, give us that $\displaystyle f(x) > 0 ~ \forall x \in (0, \frac{\pi}{2})$ and therefore $\displaystyle tan(x) > x ~ \forall x \in (0, \frac{\pi}{2})$