Let $x=e^{2\pi i/13}$. Then $$i\tan{2\pi/13}=\frac{x^2-1}{x^2+1}=\frac{x^2-x^{26}}{x^2+1}$$
(recall that $x^{13}=1$)
$$=x^2(1-x^2)(1+x^4+x^8+x^{12}+x^3+x^7)$$
$$=(x+x^2+x^5+x^6+x^9+x^{10}-x^3-x^4-x^7-x^8-x^{11}-x^{12})$$
$$4i\sin{6\pi/13}=2(x^3-x^{10})$$
So $i\tan{2\pi/13}+4i\sin{6\pi/13}=(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})$
Recall that $1+x+x^2+\cdots+x^{12}=0$.
After some tedious computation, we arrive at
$$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})$$
$$=4+x+x^3+x^4+x^9+x^{10}+x^{12}$$
The key step in the deduction is the
famous exponential sum of Gauss, which gives,
$$1+2(x+x^4+x^9+x^3+x^{12}+x^{10})=\sqrt{13}.$$
Hence $$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})=(7+\sqrt{13})/2$$
Recall our formula $1+x+x^2+\cdots+x^{12}=0$ again, and
$$(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})^2=(-1)^2-4\times(7+\sqrt{13})/2$$
$$=-13-2\sqrt{13}$$
Hence $i\tan{2\pi/13}+4i\sin{6\pi/13}=\pm i\sqrt{13+2\sqrt{13}}$
and it is obvious that $\tan{2\pi/13}+4\sin{6\pi/13}=\sqrt{13+2\sqrt{13}}$, Q.E.D.