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

$\sin{\theta} =\dfrac{1}{3}$ $\cos{\theta} =-\dfrac{2\sqrt{2}}{3}$ $\tan{\theta}=-\dfrac{\sqrt{2}}{4}$ $\sec{\theta} =-\dfrac{3\sqrt{2}}{4}$ $\cot{\theta} = -2 \sqrt{2}$
$\sin{\theta} = \dfrac{1}{\csc{\theta}} = \dfrac{1}{3}$ 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, $\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$ $\cos{\theta} = -\sqrt{1-\left(\dfrac{1}{3} \right)^2}= -\dfrac{2\sqrt{2}}{3}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{\dfrac{1}{3}}{-\dfrac{2\sqrt{2}}{3}} = -\dfrac{\sqrt{2}}{4}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\sec{\theta} = \dfrac{1}{-\dfrac{2\sqrt{2}}{3}} = -\dfrac{3\sqrt{2}}{4}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\dfrac{\sqrt{2}}{4}} = -2 \sqrt{2}$