## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sec{\theta} =-\dfrac{\sqrt{10}}{3}$ $\cos{\theta} =-\dfrac{3 \sqrt{10}}{10}$ $\sin{\theta}=\dfrac{\sqrt{10}}{10}$ $\csc{\theta} =\sqrt{10}$ $\cot{\theta} =-3$
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 $\sin{\theta}= -\dfrac{3 \sqrt{10}}{10} \times -\dfrac{1}{3} = \dfrac{\sqrt{10}}{10}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{\dfrac{\sqrt{10}}{10}}= \sqrt{10}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\dfrac{1}{3}} = -3$