Answer
$\sin{\theta} =\dfrac{1}{3}$
$\cos{\theta} =-\dfrac{2\sqrt{2}}{3}$
$\tan{\theta}=-\dfrac{\sqrt{2}}{4}$
$\sec{\theta} =-\dfrac{3\sqrt{2}}{4}$
$\cot{\theta} = -2 \sqrt{2}$
Work Step by Step
$\sin{\theta} = \dfrac{1}{\csc{\theta}} = \dfrac{1}{3}$
Since $\cot{\theta} <0$ then $\tan{\theta} <0$.
With $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} <0$ and $\sin{\theta}>0$, then $\cos{\theta} <0$.
Thus,
$\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$
$\cos{\theta} = -\sqrt{1-\left(\dfrac{1}{3} \right)^2}= -\dfrac{2\sqrt{2}}{3}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{\dfrac{1}{3}}{-\dfrac{2\sqrt{2}}{3}} = -\dfrac{\sqrt{2}}{4}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{-\dfrac{2\sqrt{2}}{3}} = -\dfrac{3\sqrt{2}}{4}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{-\dfrac{\sqrt{2}}{4}} = -2 \sqrt{2}$
