[tex]\cos\theta=\frac14[/tex] is positive, and we want [tex]\sin\theta[/tex] to be negative, which together means [tex]\theta[/tex] is an angle that lies in the fourth quadrant.
Use the Pythagorean identity to solve for [tex]\sin\theta[/tex]:
[tex]\sin^2\theta+\cos^2\theta=1\implies\sin\theta=-\sqrt{1-\cos^2\theta}=-\dfrac{\sqrt{15}}4[/tex]
Then by definition of tangent, we have
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt{15}}4}{\frac14}=-\sqrt{15}[/tex]