## Calculus (3rd Edition)

$\tan\theta= -2$
$\tan^{2}\theta=\sec^{2}\theta-1=5-1=4$ $\tan \theta=±2$ $\cos \theta>0$ (as $\sec\theta>0$) $\sin\theta<0$ (given) $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{negative\,value}{positive\,value}=negative\, value$ Therefore, $\tan\theta=-2$