Answer
$\sec{\theta} =-\dfrac{\sqrt{10}}{3}$
$\cos{\theta} =-\dfrac{3 \sqrt{10}}{10}$
$\sin{\theta}=\dfrac{\sqrt{10}}{10}$
$\csc{\theta} =\sqrt{10}$
$\cot{\theta} =-3$
Work Step by Step
Recall:
$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}}$
Since $\tan{\theta} <0 \hspace{5pt} \text{ and } \hspace{5pt} \sin{\theta}>0$, then $\cos{\theta} <0$.
With $\sec{\theta} = \dfrac{1}{\cos{\theta}} \hspace{5pt} \text{ and } \hspace{5pt} \cos{\theta} <0$, then $\sec{\theta} <0$.
Since $1+\tan^2{\theta} = \sec^2{\theta}$ and $\sec{\theta}<0$, then $\sec{\theta} = - \sqrt{1+\tan^2{\theta}}$.
Thus,
$\sec{\theta} = - \sqrt{1+\left(-\dfrac{1}{3} \right)^2} =-\dfrac{\sqrt{10}}{3}$
Since $\cos{\theta} = \dfrac{1}{\sec{\theta}}$, then
$\cos{\theta} =\dfrac{1}{-\dfrac{\sqrt{10}}{3}} = -\dfrac{3 \sqrt{10}}{10}$
With $\sin{\theta} = \cos{\theta} \tan{\theta}$, then
$\sin{\theta}= -\dfrac{3 \sqrt{10}}{10} \times -\dfrac{1}{3} = \dfrac{\sqrt{10}}{10}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{\dfrac{\sqrt{10}}{10}}= \sqrt{10}$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{-\dfrac{1}{3}} = -3$
