Tangent Inequality

From ProofWiki
Jump to navigation Jump to search


$x < \tan x$

for all $x$ in the interval $\left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.


Let $f \left({x}\right) = \tan x - x$.

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

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

Hence $f' \left({x}\right) > 0$.

From Derivative of Monotone Function, $f \left({x}\right)$ is strictly increasing in this interval.

Since $f \left({0}\right) = 0$, it follows that $f \left({x}\right) > 0$ for all $x$ in $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.