Respuesta :
[tex]\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}
\end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf 2\cdot \cfrac{17}{12}\implies \cfrac{17}{6}\qquad thus\quad\cfrac{17\pi }{12}\iff\cfrac{\frac{17\pi }{6}}{2} \\\\\\ now\quad \cfrac{17}{6}\iff 2+\cfrac{5}{6}\quad or\quad 2\pi +\cfrac{5\pi }{6} \\\\\\ \textit{so the angle goes one way around the circle }2\pi \\ \textit{and has a co-terminal angle of }\cfrac{5\pi }{6}\\\\ \textit{so, let's use that one then}[/tex]
[tex]\bf tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{sin\left( \frac{5\pi }{6} \right)}{1+cos\left( \frac{5\pi }{6} \right)}\implies tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{\frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \\\\\\ tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{\frac{1}{2}}{\frac{2-\sqrt{3}}{2}}\implies tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{1}{2}\cdot \cfrac{2}{2-\sqrt{3}} \\\\\\ tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{1}{2-\sqrt{3}}[/tex]
now, let us use the conjugate of the denominator, to rationalize it
[tex]\bf \cfrac{1}{2-\sqrt{3}}\cdot \cfrac{2+\sqrt{3}}{2+\sqrt{3}}\implies \cfrac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2}\implies \cfrac{2+\sqrt{3}}{4-3}\implies \boxed{2+\sqrt{3}}[/tex]
[tex]\bf 2\cdot \cfrac{17}{12}\implies \cfrac{17}{6}\qquad thus\quad\cfrac{17\pi }{12}\iff\cfrac{\frac{17\pi }{6}}{2} \\\\\\ now\quad \cfrac{17}{6}\iff 2+\cfrac{5}{6}\quad or\quad 2\pi +\cfrac{5\pi }{6} \\\\\\ \textit{so the angle goes one way around the circle }2\pi \\ \textit{and has a co-terminal angle of }\cfrac{5\pi }{6}\\\\ \textit{so, let's use that one then}[/tex]
[tex]\bf tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{sin\left( \frac{5\pi }{6} \right)}{1+cos\left( \frac{5\pi }{6} \right)}\implies tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{\frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \\\\\\ tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{\frac{1}{2}}{\frac{2-\sqrt{3}}{2}}\implies tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{1}{2}\cdot \cfrac{2}{2-\sqrt{3}} \\\\\\ tan\left( \cfrac{\frac{5\pi }{6}}{2} \right)=\cfrac{1}{2-\sqrt{3}}[/tex]
now, let us use the conjugate of the denominator, to rationalize it
[tex]\bf \cfrac{1}{2-\sqrt{3}}\cdot \cfrac{2+\sqrt{3}}{2+\sqrt{3}}\implies \cfrac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2}\implies \cfrac{2+\sqrt{3}}{4-3}\implies \boxed{2+\sqrt{3}}[/tex]
