Answer
$\cos{\theta} =- \dfrac{\sqrt{5}}{3}$
$\tan{\theta}=-\dfrac{2 \sqrt{5}}{5}$
$\csc{\theta} =\dfrac{3 }{2}$
$\sec{\theta} =-\dfrac{3 \sqrt{5}}{5}$
$\cot{\theta} =-\dfrac{\sqrt{5}}{2}$
Work Step by Step
$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} <0$
Since $\sin{\theta} >0$, then $\cos{\theta}$ must be negative.
Thus,
$\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$
$\cos{\theta} = -\sqrt{1-\left(\dfrac{2}{3} \right)^2}= - \dfrac{\sqrt{5}}{3}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{\dfrac{2}{3}}{- \dfrac{\sqrt{5}}{3}} = -\dfrac{2 \sqrt{5}}{5}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{\dfrac{2}{3}}= \dfrac{3 }{2}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{-\dfrac{\sqrt{5}}{3}} = -\dfrac{3 \sqrt{5}}{5}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{-\dfrac{2 \sqrt{5}}{5}} = -\dfrac{\sqrt{5}}{2}$
