Given that $\tan {x\over2} =\frac{ 1 - \cos(x)} {\sin(x)}$, deduce that $\tan {\pi\over12} = 2-\sqrt3$.
I know $\tan {x\over2} =\frac{ 1 - \cos(x)} {\sin(x)}$ is true and I can prove it by squaring and taking a square root of the right side then I multiply by $0.5\over0.5$. And I will use $$\cos{x\over2}=\sqrt{\frac{1+\cos x}{2}}$$ and $$\sin{x\over2}=\sqrt{\frac{1-\cos x}{2}}$$
But I do not understand the part of deducing.
I suspect this is what you need to solve:
Show that $\tan(\frac{1}{12}\pi)=2-\sqrt{3}$, given that $$\tan(x/2)=\frac{1-\cos(x)}{\sin(x)}.$$
This can be shown quite easily as $\cos(\frac{1}{6}\pi)=\frac{1}{2}\sqrt{3}$ and $\sin(\frac{1}{6}\pi)=\frac{1}{2}$. (These are well known values and I suspect you do not need to proof this.)