Answer
$\frac{12\sqrt {85}}{49}$
Work Step by Step
Let $sin^{-1}\frac{6}{11}=t$, we have $sin(t)=\frac{6}{11}, cos(t)=\frac{\sqrt {11^2-6^2}}{11}=\frac{\sqrt {85}}{11}$ and $tan(t)=\frac{6}{\sqrt {85}}$. Thus
$tan(2t)=\frac{2tan(t)}{1-tan^2(t)}=\frac{2(\frac{6}{\sqrt {85}})}{1-(\frac{6}{\sqrt {85}})^2}=\frac{12\sqrt {85}}{49}$
