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If $R$ and $r$ are the radii of the circumcircle and incircle of a regular polygon of $n$ sides, each side being of length a, then $2(R+r)$ equals
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The correct answer is:
$\operatorname{a~} \cot \frac{\pi}{2 n}$
We know that $R=\frac{1}{2}$ a $\operatorname{cosec}\left(\frac{\pi}{n}\right)$ and $r=\frac{1}{2}$ a $\cot \left(\frac{\pi}{n}\right)$. so that
$2(R+r)=a\left(\frac{1}{\sin \left(\frac{\pi}{n}\right)}+\frac{\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}\right)=a \frac{1+\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}$
$=a \frac{2 \cos ^2\left(\frac{\pi}{2 n}\right)}{2 \sin \left(\frac{\pi}{2 n}\right) \cos \left(\frac{\pi}{2 n}\right)}=a \cot \frac{\pi}{2 n}$
$2(R+r)=a\left(\frac{1}{\sin \left(\frac{\pi}{n}\right)}+\frac{\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}\right)=a \frac{1+\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}$
$=a \frac{2 \cos ^2\left(\frac{\pi}{2 n}\right)}{2 \sin \left(\frac{\pi}{2 n}\right) \cos \left(\frac{\pi}{2 n}\right)}=a \cot \frac{\pi}{2 n}$
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