How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$ Nice  Question:
show that: The follow nice trigonometry 

$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$

This problem I have ugly solution, maybe someone have nice methods? Thank you 
My ugly solution:

let $$A=\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}},B=\tan{\dfrac{4\pi}{13}}+4\sin{\dfrac{\pi}{13}}$$
  since
  $$\tan{w}=2[\sin{(2w)}-\sin{(4w)}+\sin{(6w)}-\sin{(8w)}+\cdots\pm \sin{(n-1)w}]$$
  where $n$ is odd,and $w=\dfrac{2k\pi}{n}$

so

$$\tan{\dfrac{2\pi}{13}}=2\left(\sin{\dfrac{4\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{6\pi}{13}}+\sin{\dfrac{2\pi}{13}}\right)$$
  $$\tan{\dfrac{4\pi}{13}}=2\left(\sin{\dfrac{5\pi}{13}}-\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{6\pi}{13}}-\sin{\dfrac{\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)$$
  then
  $$A^2-B^2=(A+B)(A-B)=16\left(\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)\left(\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{6\pi}{13}}\right)=\cdots=4\sqrt{13}$$
  $$AB=\cdots=6\left(\cos{\dfrac{\pi}{13}}+\cos{\dfrac{2\pi}{13}}+\cos{\dfrac{3\pi}{13}}-\cos{\dfrac{4\pi}{13}}-\cos{\dfrac{5\pi}{13}}+\cos{\dfrac{6\pi}{13}}\right)=\cdots=3\sqrt{3}$$
  so
  $$A=\sqrt{13+2\sqrt{13}},B=\sqrt{13-2\sqrt{13}}$$

Have other nice metods?
and I know this is simlar 1982 AMM problem: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
But My problem is hard then AMM problem。Thank you very much!

Straightforward for WA, not so by hand: Let $x = \exp(i \theta)$ be a primitive $13$th root of unity and $y = \tan(\theta) + 4 \sin(3\theta)$. Then $(x, y)$ is a common zero of the polynomials $$i (x^2-1) x^3 + 2i (x^6-1)(x^2+1) + y (x^2+1)x^3 \textrm{ and}\\
x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.$$
The resultant of these polynomials in the variable $x$ is $(y^4-26y^2+117)^3$.  The roots of this resultant are the possible values of $y$ for different choices of $x$.  Since $\sin(6 \pi/13)$ is close to one the value of $y$ in this specific case must be $\sqrt{13+2\sqrt{13}}$.
