Cunning ploy

This is a little overkill solution: let $w=e^{2 \pi i/13}$ and let $V$ be the value we want, so, if we multiply both sides by $i$ we get:

$iV=i\tan\left(\dfrac{2\pi}{13}\right)+4i\sin\left(\dfrac{6\pi}{13}\right)$

Also, we know that $w^k+1/w^k=2\cos\left(\dfrac{2\pi k}{13}\right)$ , $w^k-1/w^k=2i\sin\left(\dfrac{2\pi k}{13}\right)$ , $w^{13}=1$ , $w+w^2+\cdots+w^{11}+w^{12}=-1$ and $w^{-k}=w^{13-k}$

So, the expression becomes:

$iV=\dfrac{w-1/w}{w+1/w}+2(w^3-1/w^3) \\ iV=\dfrac{w^2-1}{w^2+1}+2w^3-2w^{10}$

With a little trick we can delete that fraction:

$iV=\dfrac{w^2(1-w^{24})}{w^2+1}+2w^3-2w^{10} \\ iV=\dfrac{w^2(1+w^2)(1-w^2)(1+w^4)(1+w^8+w^{16})}{1+w^2}+2w^3-2w^{10}$

After expanding and simplifying we get:

$iV=w+w^2+w^3+w^5+w^6+w^9-(w^4+w^7+w^8+w^{10}+w^{11}+w^{12})$

Now, let $a=w+w^2+w^3+w^5+w^6+w^9$ and $b=w^4+w^7+w^8+w^{10}+w^{11}+w^{12}$ , so $iV=a-b$

Try to find $a+b$ and $ab$ , the first one is easy:

$a+b=w+w^2+\cdots+w^{11}+w^{12}=-1$

Now, to find $ab$ we need to expand a large product, which after simplification is:

$ab=4+w+w^3+w^4+w^9+w^{10}+w^{12}$

From here, let $c=w+w^3+w^4+w^9+w^{10}+w^{12}$ and $d=w^2+w^5+w^6+w^7+w^8+w^{11}$ , so we can find $c+d$ and $cd$ again:

$c+d=w+w^2+\cdots+w^{11}+w^{12}=-1$

The product $cd$ is also large, which after simplificacion is:

$cd=-3$

That is great, because we can get $c$ , and then $a-b$ :

$(c-d)^2=(c+d)^2-4cd=(-1)^2-4(-3)=13$

Since $c>0$ and $d<0$ , then $c-d=\sqrt{13}$ .

So, $c=\dfrac{-1+\sqrt{13}}{2}$ (earlier I also posted this part of the problem as another problem here ).

Now, let's obtain $ab$ :

$ab=4+\dfrac{-1+\sqrt{13}}{2}=\dfrac{7+\sqrt{13}}{2}$

Finally, obtain $V$ , with the fact that $V>0$ :

$(a-b)^2=(a+b)^2-4ab=(-1)^2-2(7+\sqrt{13})=-(13+2\sqrt{13}) \\ a-b=\sqrt{13+2\sqrt{13}}i\\ iV=\sqrt{13+2\sqrt{13}}i$

Finally, $V=\sqrt{13+2\sqrt{13}}$

On comparing we get $A=B=13$ , thus the answer is $A+B=\boxed{26}$ .