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 $t$.
 A: Note that $$\sin^{2}\left(\frac{x}{2}\right) = \frac{1-\cos x}{2},\quad \cos^{2}\left(\frac{x}{2}\right) = \frac{1+\cos x}{2},$$
so $$\tan^{2}\left(\frac{x}{2}\right) =\frac{1-\cos x}{1+\cos x}.$$
Setting $$t = \tan \left(\frac{x}{2}\right)$$
we get
\begin{align*}
t^{2} &= \frac{1-\cos x}{1+\cos x}\\ \implies t^{2} + t^{2}\cos x &= 1 - \cos x\\ \implies (1+t^{2})\cos x &= 1-t^{2}\\
\implies \cos x &= \frac{1-t^{2}}{1+t^{2}}
\end{align*}
*Edited based on Arnaldo's suggestion.
A: Following your idea you can do:
$$\tan x=\frac{2\tan (x/2)}{1-\tan^2(x/2)}=\frac{2t}{1-t^2}$$
so,
$$\frac{\sin x}{\cos x}=\frac{2t}{1-t^2}\to \frac{\sin^2 x}{\cos^2 x}=\frac{4t^2}{(1-t^2)^2}$$
using proportion properties we get:
$$\frac{\cos^2 x}{\sin^2 x+\cos^2 x}=\frac{(1-t^2)^2}{4t^2+(1-t^2)^2}\to \cos^2x=\frac{(1-t^2)^2}{(1+t^2)^2}\to \cos x=\frac{1-t^2}{1+t^2}$$
A: Since you are given four choices, all you have to do is eliminate the wrong ones. If $x=0$, then $t=0$ and $\cos(x)=1$. The formulas in b) and d) give $0$ which is wrong. If $x=\pi/2$, then $t=1$ and the formula in a) gives division by $0$. The only choice left is c).
A: 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 $\cos(x)=\frac{2}{{t^2+1}}-1=\frac{1-t^2}{1+t^2}$.
