Note that the trig expression takes on the form $\displaystyle \cos(2\vartheta)$, where $\displaystyle \vartheta=\tan^{-1}x$
Recall that $\displaystyle \cos(2\vartheta)=1-2\sin^2(\vartheta)$
Now let us set up a triangle that represents the angle $\displaystyle \vartheta=\tan^{-1}x$.
We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:
$\displaystyle x^2+1=h^2\implies h=\sqrt{1+x^2}$
Now, let us evaluate the expression:
$\displaystyle 1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2$
$\displaystyle \sin\vartheta$ can be determined from the triangle we constructed.
Thus, $\displaystyle \sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2 }}$
Substitute this into the equation and simplify:
$\displaystyle 1-2(\sin\vartheta)^2=1-2\left(\frac{x}{\sqrt{1+x^2}}\right)^2=\dots$
I'll leave the rest for you to do.
I hope this makes sense!
--Chris