Tangent Inequality

Theorem

$x < \tan x$

for all $x$ in the interval $\openint 0 {\dfrac \pi 2}$.

Proof

Let $\map f x = \tan x - x$.

By Derivative of Tangent Function, $\map {f'} x = \sec^2 x - 1$.

By Shape of Secant Function, $\sec^2 x > 1$ for $x \in \openint 0 {\dfrac \pi 2}$.

Hence $\map {f'} x > 0$.

From Derivative of Monotone Function, $\map f x$ is strictly increasing in this interval.

Since $\map f 0 = 0$, it follows that $\map f x > 0$ for all $x \in \openint 0 {\dfrac \pi 2}$.

$\blacksquare$