# Given that \tan \frac x2=\frac{1-\cos(x)}{sin(x)}, deduce that \tan \frac{\pi}{12}=2−\sqrt3

Given that $\mathrm{tan}\frac{x}{2}=\frac{1-\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$, deduce that $\mathrm{tan}\frac{\pi }{12}=2-\sqrt{3}$
I know $\mathrm{tan}\frac{x}{2}=\frac{1-\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$ is true and I can prove it by squaring and taking a square root of the right side then I multiply by $\frac{0.5}{0.5}$. And I will use
$\mathrm{cos}\frac{x}{2}=\sqrt{\frac{1+\mathrm{cos}x}{2}}$
and
$\mathrm{sin}\frac{x}{2}=\sqrt{\frac{1-\mathrm{cos}x}{2}}$
But I do not understand the part of deducing.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

I suspect this is what you need to solve:
Show that $\mathrm{tan}\left(\frac{1}{12}\pi \right)=2-\sqrt{3}$, given that
$\mathrm{tan}\left(\frac{x}{2}\right)=\frac{1-\mathrm{cos}x}{\mathrm{sin}x}$
This can be shown quite easily as $\mathrm{cos}\left(\frac{16}{\pi }\right)=\frac{12}{\sqrt{3}}$ and $\mathrm{sin}\left(\frac{16}{\pi }\right)=\frac{12}{}$. (These are well known values and I suspect you do not need to proof this.)