<s>[INST] 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 $t$.
 [/INST] 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:
$$\frac{\cos^2 x}{\sin^2 x+\cos^2 x}=\frac{(1-t^2)^2}{4t^2+(1-t^2)^2}\to \cos^2x=\frac{(1-t^2)^2}{(1+t^2)^2}\to \cos x=\frac{1-t^2}{1+t^2}$$
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