Find the vlues of the trigonometric functions from the given information 
$\tan t = \dfrac{1}{4}$, terminal point of $t$ is in the third quadrant.

So $\tan t = \dfrac{-\sqrt{1-\cos^2 t}}{\cos t} = \dfrac{1}{4} $, because it's in the third quadrant. After solving for $\cos^2 t$ get as an answer that $\cos ^2 t = \dfrac{16}{17}$, but apparently this is wrong. Where lies my mistake? 

If 
$$\tan t = \frac{1}{4}$$
then
$$t = \arctan \left(\frac{1}{4}\right) = 14.03°$$
or $0.244$ radians.
Beware
$$\tan t = \frac{\sin t}{\cos t} = \frac{\sqrt{1 - \cos^2t}}{\cos t}$$
In your formula, there is a minus sign which is wrong.
Hence
$$\frac{\sqrt{1 - \cos^2t}}{\cos t} = \frac{1}{4}$$
$$4\sqrt{1 - \cos^2 t} = \cos t$$
Squaring
$$16 - 16\cos^2 t = \cos^2 t$$
$$16 = 17\cos^2 t$$
$$\cos t = \pm \sqrt{\frac{16}{17}} = \pm 4\sqrt{\frac{1}{17}}$$
Then choose the right sign for what you know.
