Answer
$\frac{48+25\sqrt 3}{39}$
Work Step by Step
Step 1. Letting $sin^{-1}(\frac{3}{5})=u$, we have $sin(u)=\frac{3}{5}$, thus $tan(u)=\frac{3}{4}$
Step 2. $tan(u+\frac{\pi}{6})=\frac{tan(u)+tan(\frac{\pi}{6})}{1-tan(u)tan(\frac{\pi}{6})}=\frac{\frac{3}{4}+\frac{\sqrt 3}{3}}{1-(\frac{3}{4})(\frac{\sqrt 3}{3})}=\frac{9+4\sqrt 3}{3(4-\sqrt 3)}\times\frac{4+\sqrt 3}{4+\sqrt 3}=\frac{36+12+16\sqrt 3+9\sqrt 3}{39}=\frac{48+25\sqrt 3}{39}$
