## Precalculus (6th Edition)

$-\sqrt {\dfrac {7}{11}}$
$\tan \theta =-\dfrac {\sqrt {7}}{2}$ (negative) $sec\theta =\dfrac {1}{\cos \theta } > 0$ ( positive) So the angle is located in quadrant 4 (sine function is negative in quadrant 4) $\Rightarrow \sin \theta =\dfrac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}=\dfrac {\dfrac {-\sqrt {7}}{2}}{\sqrt {1+\left( -\dfrac {\sqrt {7}}{2}\right) ^{2}}}=-\sqrt {\dfrac {7}{11}}$