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 :
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$
 A: 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: 


*

*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 $$\prod\limits_{k=1}^{n} \tan\Bigl(\frac{k\pi}{2n+1}\Bigr) = \frac{\sqrt{2n+1}}{2^{n}} \times \frac{2^{n-1}}{\sqrt{n}} = \frac{\sqrt{2n+1}}{2 \cdot \sqrt{n}}$$
A: I would like to address a question nobody seems to have asked, but which should be on everybody's mind:

What's the deal with the specific values $2, 5, 6$?

Or perhaps:

How could we have anticipated that an identity like this might be true, and how might we generate others like it?

This can be explained with a little Galois theory. Let $\zeta = \exp \left( \frac{2\pi i}{13} \right)$ be a primitive $13^{th}$ root of unity and let
$$a = \frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}} = i \tan \frac{2\pi}{13} \in \mathbb{Q}(\zeta).$$
Since $13$ is prime, the Galois group $(\mathbb{Z}/13\mathbb{Z})^{\ast}$ of $\mathbb{Q}(\zeta)$ is cyclic of order $12$ on generator $\sigma : \zeta \mapsto \zeta^2$. This group has a unique quotient $C_4$ of order $4$, hence by the fundamental theorem of Galois theory, the unique subextension of degree $4$ is the fixed field $K$ of $\tau = \sigma^4 : \zeta \mapsto \zeta^3$. In particular, 
$$z = a \tau(a) \tau^2(a) = \left( i \tan \frac{2\pi}{13} \right) \left( i \tan \frac{6 \pi}{13} \right) \left( i \tan \frac{5 \pi}{13} \right) \in K.$$
The claim is that this is equal to $-i \sqrt{ 65 + 18 \sqrt{13} }$, which also generates an extension of degree $4$. Well-known results about Gauss sums (equivalently, well-known results about ramification of primes in number fields) imply that the unique quadratic subextension of $\mathbb{Q}(\zeta)$, which is necessarily a subextension of $K$, is $F = \mathbb{Q}(\sqrt{13})$, of which $K$ is itself a quadratic extension. 
This suggests that we make use of the trace and norm maps $\text{Tr}_{K/F}, N_{K/F}$. In fact, $\text{Gal}(K/F)$ is generated by the image of $\mu = \sigma^2 : \zeta \mapsto \zeta^4$, so 
$$\text{Tr}_{K/F}(z) = \left( i \tan \frac{2\pi}{13} \right) \left( i \tan \frac{6 \pi}{13} \right) \left( i \tan \frac{5 \pi}{13} \right) + \left( i \tan \frac{8\pi}{13} \right) \left( i \tan \frac{11 \pi}{13} \right) \left( i \tan \frac{7 \pi}{13} \right).$$
But using the identity $\tan x = - \tan(\pi - x)$ this is clearly equal to zero. It follows that $z$ satisfies a quadratic equation of the form
$$z^2 - N_{K/F}(z) = 0$$
over $F$, so (by checking signs) it suffices to show that $N_{K/F}(z) = -65 - 18 \sqrt{13}$. 
Unfortunately, I'm out of clever ideas for this last step. But at least we now know that $N_{K/F}(z) \in F$ must be of the form $a + b \sqrt{13}$, and therefore that some identity of the desired form must be true, based on a conceptual argument rather than ad hoc computations. 

The above answers the question "given the LHS, why should we expect it to equal something like the RHS?" We can also ask "given the RHS, why should we expect it to equal something like the LHS?" The answer is the Kronecker-Weber theorem. It is a straightforward exercise to show that the number field generated by the RHS has Galois group $C_4$, which is in particular abelian, so by Kronecker-Weber the RHS lies in $\mathbb{Q}(\zeta_n)$ for some $n$ such that $C_4$ is a quotient of the Galois group $(\mathbb{Z}/n\mathbb{Z})^{\ast}$. Moreover, $\sqrt{13}$ must lie in $\mathbb{Q}(\zeta_n)$, and $n = 13$ is the smallest value of $n$ which satisfies both requirements. For the problem as written I think we actually need $n = 52$, or maybe $n = 26$, but you get the idea. 
A: HINT: The number $\alpha=\sqrt{65+18\sqrt{13}}$ is one of the four solutions of the equation $(\alpha^2-65)^2-4212=0$. Then, show that the LHS is also a solution of the same equation. The other three solutions must be disregarded somehow.
