First of all find [tex]sin \frac{17 \pi }{ 6} =sin(3 \pi - \frac{ \pi }{ 6} )= \frac{1}{2} [/tex][tex] cos \frac{17 \pi }{ 6} =cos(3 \pi - \frac{ \pi }{ 6} )= \frac{-\sqrt3}{2} [/tex]
Then use [tex]tan \frac{x}{2} = \frac{sinx}{1+cosx} [/tex] to verify [tex]tan \frac{17 \pi }{12} = \frac{1}{2- \sqrt{3} } [/tex]
