Answer
$- \sqrt {15}$
Work Step by Step
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 $\tan \theta =\tan [\cos^{-1} (-\dfrac{1}{4})]=- \sqrt {15}$
