Answer
$3-2\sqrt 2$
Work Step by Step
Given $y$ in Quadrant I and $csc(y)=3$, we have $sin(y)=\frac{1}{3}$ and $cos(y)=\sqrt {1-sin^2y}=\frac{2\sqrt 2}{3}$
Use the Half-Angle Formula, $tan\frac{y}{2}=\sqrt {\frac{1-cos(y)}{1+cos(y)}}
=\sqrt {\frac{1-\frac{2\sqrt 2}{3}}{1+\frac{2\sqrt 2}{3}}}
=\sqrt {\frac{3-2\sqrt 2}{3+2\sqrt 2}}
=\sqrt {\frac{(3-2\sqrt 2)^2}{(3+2\sqrt 2)(3-2\sqrt 2)}}=3-2\sqrt 2$
