#### Answer

$csc\theta =\dfrac {3}{2};$
$\cos \theta =-\dfrac {\sqrt {5}}{3};$
$sec\theta =-\dfrac {3}{\sqrt {5}};$
$\tan \theta =-\dfrac {2}{\sqrt {5}};$
$\cot \theta =-\dfrac {\sqrt {5}}{2}$

#### Work Step by Step

In quadrant 2 only $\csc \theta$ and $\sin\theta$ are positive other 4 trigonometric functions are negative . Then
$csc\theta =\dfrac {1}{\sin \theta }=\dfrac {\dfrac {1}{2}}{3}=\dfrac {3}{2};$
$\cos \theta =-\sqrt {1-\sin ^{2}\theta =}-\sqrt {1-\left( \dfrac {2}{3}\right) ^{2}}=-\dfrac {\sqrt {5}}{3};$
$sec\theta =\dfrac {1}{\cos \theta }=\dfrac {1}{-\dfrac {\sqrt {5}}{3}}=-\dfrac {3}{\sqrt {5}};$
$\tan \theta =\dfrac {\sin \theta }{\cos \theta }=\dfrac {\dfrac {2}{3}}{-\dfrac {\sqrt {5}}{3}}=-\dfrac {2}{\sqrt {5}};$
$\cot \theta =\dfrac {1}{\tan \theta }=-\dfrac {\sqrt {5}}{2}$