## Precalculus (6th Edition)

$-\dfrac {\sqrt {45}}{7}$
$sec\theta =\dfrac {1}{\cos \theta }=\dfrac {7}{2}$ (positive) $\tan \theta < 0$ (negative) So angle is located in quadrant 4 ( sine is negative in quadrant 4) $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {2}{7}\right) ^{2}}=-\dfrac {\sqrt {45}}{7}$