$\cos(\tan^{-1}(x))$
I set $\tan^{-1}(x)=\theta$, then I found the inverse, which is $\tan(\theta) = x$. That would mean that $\tan(\theta) = \frac{opp}{adj} = \frac{x}{1}$. What I need to do is find the hypotenuse. I know I have to use the Pythagorean theorem, but I don't have the hypotenuse, so it's not possible.
Since you know you have to use Pythagoras' theorem, by the converse of Pythagoras' theorem, the triangle must be a right triangle. Thus, since $opp.=x,adj.=1$, drawing out the triangle (if you need to), and using Pythagoras' theorem gives $$\boxed{Hypotenuse=\sqrt{x^2+1}}$$