## Precalculus (10th Edition)

$\sin\theta=-\dfrac{\sqrt 5}{3}$ $\tan\theta=-\dfrac{\sqrt 5}{2}$ $\cot\theta=-\dfrac{2\sqrt 5}{5}$ $\sec\theta=\dfrac{3}{2}$ $\csc\theta=-\dfrac{3\sqrt 5}{5}$
Determine $\sin\theta$: $\sin^2\theta+\cos^2\theta=1$ $\sin\theta=\pm\sqrt{1-\cos^2\theta}=\pm\sqrt{1-\left(\dfrac{2}{3}\right)^2}$ $=\pm\sqrt{\dfrac{5}{9}}$ $=\pm\dfrac{\sqrt 5}{3}$ Because the angle $\theta$ is in Quadrant IV, we have: $\sin\theta=-\dfrac{\sqrt 5}{3}$ Determine $\tan\theta,\cot\theta,\sec\theta,\csc\theta$: $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt 5}{3}}{\frac{2}{3}}=-\dfrac{\sqrt 5}{2}$ $\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{-\frac{\sqrt 5}{2}}=-\dfrac{2}{\sqrt 5}=-\dfrac{2\sqrt 5}{5}$ $\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{1}{\frac{2}{3}}=\dfrac{3}{2}$ $\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{-\frac{\sqrt 5}{3}}=-\dfrac{3}{\sqrt 5}=-\dfrac{3\sqrt 5}{5}$