The fraction $\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}$ is equal to 
$\textbf{(A)}\ \frac{2\sqrt2}{3} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{2\sqrt3}3 \qquad \textbf{(D)}\ \frac43 \qquad \textbf{(E)}\ \frac{16}{9}$
Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields
\[\sqrt{\left(\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}\right)^2}\]
\[\sqrt{\frac{4(2+4\sqrt{3}+6)}{9(2+\sqrt{3})}}\]
\[\sqrt{\frac{4(8+4\sqrt{3})}{9(2+\sqrt{3})}}\]
\[\sqrt{\frac{16}{9}}\]
\[\frac43\]
The answer is $\boxed{\textbf{(D)}}$.