Answer
$tan~(2~cos^{-1}~\frac{1}{4}) = -\frac{\sqrt{15}}{7}$
Work Step by Step
Let $~~\theta = cos^{-1}~\frac{1}{4}$
Then: $~~cos~\theta = \frac{1}{4}$
Then: $~~sin~\theta = \frac{\sqrt{4^2-1^2}}{4} = \frac{\sqrt{15}}{4}$
We need to find $tan~2\theta$:
$tan~2\theta = \frac{sin~2\theta}{cos~2\theta}$
$tan~2\theta = \frac{2~sin~\theta~cos~\theta}{cos^2~\theta-sin^2~\theta}$
$tan~2\theta = \frac{(2)(\frac{\sqrt{15}}{4})(\frac{1}{4})}{(\frac{1}{4})^2-(\frac{\sqrt{15}}{4})^2}$
$tan~2\theta = \frac{(2)(\frac{\sqrt{15}}{4})(\frac{1}{4})}{(\frac{1}{4})^2-(\frac{\sqrt{15}}{4})^2}$
$tan~2\theta = \frac{\frac{2\sqrt{15}}{16}}{(\frac{1}{16})-(\frac{15}{16})}$
$tan~2\theta = \frac{\frac{2\sqrt{15}}{16}}{-\frac{14}{16}}$
$tan~2\theta = -\frac{2\sqrt{15}}{14}$
$tan~2\theta = -\frac{\sqrt{15}}{7}$
Therefore, $~~tan~(2~cos^{-1}~\frac{1}{4}) = -\frac{\sqrt{15}}{7}$
