Given cos(75∘)=6−24\displaystyle \cos (75^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}cos(75∘)=46−2 and sin(75∘)=6+24\displaystyle \sin (75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}sin(75∘)=46+2
Show that tan2(75∘)\tan^2 (75^\circ)tan2(75∘) can be written in the form a+b3a + b\sqrt{3}a+b3
Fully justify your answer.