## Algebra 2 (1st Edition)

Double -angle Theorem can be defined as: $\tan \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{1+ \cos \theta}}$ $\tan (15^{\circ})=\tan \dfrac{(30^{\circ})}{2}$ or, $=\sqrt {\dfrac{1- \cos (30^{\circ})}{1+ \cos (30^{\circ})}}$ or, $=\sqrt {\dfrac{1- \cos (\sqrt 3/2)}{1+ \cos (\sqrt 3/2)}}$ or, $=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}}$ or, $=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}} \times \sqrt {\dfrac{2-\sqrt 3}{2-\sqrt 3}}$ or, $=2-\sqrt 3$ Hence, we can see that only option $B$ satisfies the equation.