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For any acute angle $\alpha$, then $\tan \alpha \geq \alpha$. Is it possible to prove this geometrically?

My work so far:

A rigorous proof requires defining a radian measure of angles: this can be done in one of two ways:

  1. Arclength. E.g. define a semicircle to be $\pi$, and the measure of angle to be the portion of the semicircle it spans.
  2. Area. E.g. define the area of a circle to be $\pi$, and the measure of angle to be its area compared to the area of a semicircle.

Using arclength seems simpler (although area may end up being more rigorous), so I'll start with that.

There is a similar identity $\sin \alpha \leq \alpha$. This follows directly from the triangle inequality: Picture from https://www2.clarku.edu/faculty/djoyce/trig/sines.html

$\overset\frown {BD} \geq BD \geq BC$.

I tried to do something with the tangent segment (not show on the diagram, but the segment from $D$ parallel to $BC$ until intersecting line $AB$), but was unable to prove this.

SRobertJames
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    See for example my answer here. – dxiv Jun 16 '23 at 01:10
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    This answer might be of some help. Note inequality $(1)$. – robjohn Jun 16 '23 at 01:10
  • Continue from your attempt "the segment from $D$ parallel to $BC$ until intersecting line $AB$". The area between such segment and centre $A$ is a right-angled triangle, and this right-angled triangle has a larger area than the sector $A\overset\frown {BD}$. – peterwhy Jun 16 '23 at 01:42
  • @peterwhy Yes, that seems to work, thank you. It uses of course the area definition. Is there a way to prove this using the arclength definition (which seems more natural and basic to me)? – SRobertJames Jun 16 '23 at 01:48
  • @SRobertJames FWIW the link in my first comment uses the arclength definition. – dxiv Jun 16 '23 at 02:07
  • Draw your tangent segment at $D$ to intersect $AB$ in $F$. Draw another tangent at $B$ to intersect $DF$ in $G$. The length $BF$ (ie $\tan\alpha$) is greater than sum of lengths of $BG$ and $DG$ and this sum is greater than length of arc (ie $\alpha$). See https://math.stackexchange.com/q/1785841/72031 – Paramanand Singh Jun 29 '23 at 04:16

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