# Prove $4\sin(2\pi/5)+\tan(2\pi/5)=5\cot(\pi/5)$

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.

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 $$\angle ACD=\angle ADC=\dfrac{2\pi}{5}$$.