<s>[INST] Showing $\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}$ How can one show :
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$
 [/INST] I delete my comments and sum them up as a possible method which can lead to a solution.
Here are my attempts:
$\textbf{Attempt 1}$. Observe that if: $A+B+C = \pi$, then $$ \tan(\pi - A) = \tan(B+C) = \frac{\tan{B}+\tan{C}}{1-\tan{B} \cdot \tan{C}}$$ So you get $$\tan{A} + \tan{B} + \tan{C} = \tan{A} \cdot \tan{B} \cdot \tan{C}$$ So now you want to find the value of $$\tan\frac{2\pi}{13} + \tan\frac{5\pi}{13} + \tan\frac{6 \pi}{13}$$ which  $\mathsf{I \ don't \ know}$. You want to see this post: 


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*Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$
As I say in the post you have to look for some equations for which $\tan$ appears as a root and then perhaps look at sum of the roots. 

$\textbf{Attempt 2.}$ Consider the equation $x^{2n}-1=0$. The roots of the equation are, $$1,-1, \cos\frac{\pi}{n}+i\cdot \sin\frac{\pi}{n},\cdots, \cos\frac{(2n-1)\pi}{n}+ i \cdot \sin\frac{(2n-1)\pi}{n}$$ Therefore we can write $$ \small x^{2n}-1 = (x-1) \cdot (x+1) \cdot \biggl(x -\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr) \cdots \biggl(x - \cos\frac{(n-1)\pi}{n}-i\sin\frac{(n-1)\pi}{n}\biggr)$$ Also we have $$\cos\frac{(2n-k)\pi}{n} = \cos\frac{k\pi}{n} \ \text{and} \ \sin\frac{(2n-k)\pi}{n} =  -\sin\frac{k\pi}{n}$$
Using this we can pair up $$\biggl(x-\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr)\cdot \biggl(x-\cos\frac{(2n-1)\pi}{n}-i\sin\frac{(2n-1)\pi}{n}\biggr)$$ $$ = x^{2} - 2\cdot x \cdot \cos\frac{\pi}{n} + 1$$
Similarly $$\biggl(x-\cos\frac{2\pi}{n}-i\sin\frac{2\pi}{n}\biggr) \cdot (x - \cos\frac{(2n-2)\pi}{n}-i \sin\frac{(2n-2)\pi}{n}$$ $$= x^{2} -2 \cdot x \cdot \cos\frac{2\pi}{n} + 1$$
Continuing in this way, we get 
\begin{align*}
\frac{x^{2n}-1}{x^{2}-1} &= 1 + x^{2} + x^{4} + \cdots + x^{2n-2} \\\ &= \small\biggl(x^{2} - 2x \cos\frac{\pi}{n}+1\biggr) \cdot \biggl(x^{2}-2x\cos\frac{2\pi}{n}+1\biggr) \cdots \biggl(x^{2} - 2x \cos\frac{(n-1)\pi}{n}+1\biggr)
\end{align*}
Now putting $x=1$ and using the fact that $1-\cos{x} = 2 \sin^{2}\frac{x}{2}$ we get $$n = 4^{n-1} \cdot \sin^{2}\Bigl(\frac{\pi}{2n}\Bigr) \cdot \sin^{2}\Bigl(\frac{2\pi}{2n}\Bigr) \cdots \sin^{2}\Bigl(\frac{(n-1)\pi}{2n}\Bigr)$$ From this we get $$\prod\limits_{k=1}^{n-1} \sin\biggl(\frac{\pi k}{2n}\biggr) = \frac{\sqrt{n}}{2^{n-1}}$$
Following the similar procedure gives $$ \prod\limits_{k=1}^{n}\sin\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{2n+1}}{{2^n}}, \qquad \prod\limits_{k=1}^{n}\cos\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{n}}{{2^{n-1}}}$$
On dividing the above you get $$\prod\limits_{k=1}^{n} \tan\Bigl(\frac{k\pi}{2n+1}\Bigr) = \frac{\sqrt{2n+1}}{2^{n}} \times \frac{2^{n-1}}{\sqrt{n}} = \frac{\sqrt{2n+1}}{2 \cdot \sqrt{n}}$$
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