Answer
$\frac{2}{3}(\sqrt 5+\sqrt 2)$
Work Step by Step
Given $x,y$ in Quadrant I and $sec(x)=\frac{3}{2}, csc(y)=3$, use the Pythagorean Identities, we have
$tan^2x+1=sec^2x=\frac{9}{4}$ or $tan^2x=\frac{5}{4}, tan(x)=\frac{\sqrt 5}{2}$, and
$1+cot^2y=csc^2y=9$ or $cot^2y=8, tan(y)=\sqrt {\frac{1}{8}}=\frac{\sqrt 2}{4}$.
Thus $tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}
=\frac{\frac{\sqrt 5}{2}+\frac{\sqrt 2}{4}}{1-\frac{\sqrt 5}{2}\times \frac{\sqrt 2}{4}}
=\frac{4\sqrt 5+2\sqrt 2}{8-\sqrt {10}}
=\frac{2(2\sqrt 5+\sqrt 2)(8+\sqrt {10})}{(8-\sqrt {10})(8+\sqrt {10})}
=\frac{2}{64-10}(16\sqrt 5+10\sqrt 2+8\sqrt 2+2\sqrt 5)=\frac{2}{54}(18\sqrt 5+18\sqrt 2)
=\frac{2}{3}(\sqrt 5+\sqrt 2)$
