If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$ 
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $\tan{x}$'s and I don't know how to go to $t$, let alone express $\cos{x}$ in $t$.

In my opinion the simplest solution is as follows:
$\tan(\frac{x}{2})=t\implies\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}$;
this can be seen by constructing a triangle with opposite side of angle $\frac{x}{2}$ equal to $t$, and adjacent side equal to $1$.  
Now, it is known $\cos(x)=2\cos^2(\frac{x}{2})-1$.
$\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}\implies\cos^2(\frac{x}{2})=\frac{1}{{t^2+1}}$.
Thus $\cos(x)=\frac{2}{{t^2+1}}-1=\frac{1-t^2}{1+t^2}$.
