## Precalculus (6th Edition) Blitzer

$- \sqrt {15}$
Suppose $\theta =\cos^{-1} (-\dfrac{1}{4})$ This gives: $\cos \theta=-\dfrac{1}{3}$ Since, $r=\sqrt {x^2+y^2}$ This implies that $y=\sqrt {r^2-x^2}=\sqrt {4^2-(1)^2}= \sqrt {15}$ Therefore, we have $\tan \theta =\tan [\cos^{-1} (-\dfrac{1}{4})]=- \sqrt {15}$