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Use the area of a regular pentagon to prove that $4\sin(2\pi/5)+\tan(2\pi/5)=5\cot(\pi/5)$

Attempt: Let the side length be 1. Then by breaking down the pentagon into 5 congruent triangles, it is easy to show that the area is $A = \frac{5}{4}\cot(\pi/5) = \frac{5}{2} R^2 \sin(2\pi/5) = 5r^2 \tan(\pi/5)$ where $R$ is the circumradius and $r$ is the inradius.

Then it suffices to prove that the area can also be written as $\sin(2\pi/5)+\frac{1}{4}\tan(2\pi/5)$. I tried to break down the area of a regular pentagon as a trapezoid and triangle, but the algebra got messy: $A = (1+\cos(2\pi/5))\sin(2\pi/5)+(1/2)(1+2\cos(2\pi/5))(R+r-\sin(2\pi/5))$. I appreciate any suggestions.

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If you cut the pentagon $ABCDE$ into three triangles $ABC$, $ACD$, $ADE$ then $$ \operatorname{Area}\triangle ABC = \frac12\sin\frac{3\pi}{5}=\frac12\sin\frac{2\pi}{5} $$ and $\triangle ABC\cong\triangle AED$, so it remains to show $\operatorname{Area}\triangle ACD=\dfrac14\tan\dfrac{2\pi}{5}$, or equivalently, the perpendicular distance of $A$ from $CD$ is $\dfrac12\tan\dfrac{2\pi}{5}$ which is immediate by looking at the base angles $\angle ACD=\angle ADC=\dfrac{2\pi}{5}$.

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