I'm looking for a simple geometrical method of proving that $$\theta < \tan\theta$$ for $0 < \theta < \frac{\pi}{2}$.
I am able to prove that $\sin\theta < \theta$.
$\endgroup$ 61 Answer
$\begingroup$$${\displaystyle S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}} \tag1$$where $ {\displaystyle S_{sectKOA}}$ — area of sector ${\displaystyle KOA} $
$${\displaystyle S_{\triangle KOA}={\frac {1}{2}}\cdot |OA|\cdot |KH|={\frac {1}{2}}\cdot |OA|\cdot |OK|\cdot \sin x={\frac {1}{2}}\cdot 1\cdot 1\cdot \sin x={\frac {\sin x}{2}}}$$$${\displaystyle S_{sectKOA}={\frac {1}{2}}R^{2}x={\frac {x}{2}}}$$$${\displaystyle S_{\triangle OAL}={\frac {1}{2}}\cdot |OA|\cdot |LA|={\frac {\mathrm {tan} \,x}{2}}}$$from $\triangle OAL: |LA|={\mathrm {tan}}\,x$
substitute in $(1)$:
$$ {\frac {\sin x}{2}}<{\frac {x}{2}}<{\frac {{\mathrm {tan}}\,x}{2}}$$
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