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