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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 $\sin\theta < \theta$.

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1 Answer

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enter image description here $${\displaystyle S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}} \tag1$$ 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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