## Precalculus (6th Edition)

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