Answer
$\frac{12\sqrt {85}}{49}$
Work Step by Step
Letting $sin^{-1}\frac{6}{11}=t, 0\lt t\lt\frac{\pi}{2}$, we have $sin(t)=\frac{6}{11}, cos(t)=\frac{\sqrt {11^2-6^2}}{11}=\frac{\sqrt {85}}{11}, tan(t)=\frac{6\sqrt {85}}{85}$
Thus $tan(2t)=\frac{2tan(t)}{1-tan^2(t)}=\frac{2(\frac{6\sqrt {85}}{85})}{1-(\frac{6\sqrt {85}}{85})^2}=\frac{12\sqrt {85}}{85-36}=\frac{12\sqrt {85}}{49}$
