[tex]\textit{Half-Angle Identities}\qquad tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}~~\leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\\\ 2\cdot \cfrac{\pi }{12}\implies \cfrac{\pi }{6}~\hspace{10em} \cfrac{~~ \frac{\pi }{6}~~}{2}\implies \cfrac{\pi }{6}\cdot \cfrac{1}{2}\implies \cfrac{\pi }{12}[/tex]
so let's use the double of the angle in the half-angle identities, check the picture below for the cosine, sine pairs
[tex]tan\left( \cfrac{~~ \frac{\pi }{6}~~}{2} \right)=\cfrac{sin\left( \frac{\pi }{6} \right)}{1+cos\left( \frac{\pi }{6} \right)}\implies tan\left( \cfrac{~~ \frac{\pi }{6}~~}{2} \right)=\cfrac{~~ \frac{1}{2}~~}{1+\frac{\sqrt{3}}{2}}[/tex]
[tex]tan\left( \cfrac{~~ \frac{\pi }{6}~~}{2} \right)=\cfrac{~~ \frac{1}{2}~~}{~~\frac{2+\sqrt{3}}{2}~~}\implies tan\left( \cfrac{~~ \frac{\pi }{6}~~}{2} \right)=\cfrac{1}{2}\cdot \cfrac{2}{2+\sqrt{3}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill tan\left( \cfrac{~~ \frac{\pi }{6}~~}{2} \right)=\cfrac{1}{2+\sqrt{3}}~\hfill[/tex]
