Вычислите : tg15°(1+sin60°)

Решите задачу:

$tg15^\circ\cdot (1+sin60^\circ )=(2-\sqrt3)\cdot (1-\frac{\sqrt3}{2})=\\\\=(2-\sqrt3)\cdot \frac{2-\sqrt3}{2}=\frac{1}{2}\cdot (2-\sqrt3)^2= \frac{1}{2}\cdot (7-4\, \sqrt3) =3,5-2\sqrt3\\\\\\tg15^\circ= tg(45^\circ-30^\circ)=\frac{tg45^\cic-tg30^\circ}{1+tg45^\circ \cdot tg30^\circ } = \frac{1-\frac{\sqrt3}{3} }{1+\frac{\sqrt3}{3}} =\frac{3-\sqrt3}{3+\sqrt3}= \frac{\sqrt3-1}{\sqrt3+1}=\\\\= \frac{(\sqrt3-1)^2}{3-1}=\frac{3-2\sqrt3+1}2}= \frac{4-2\sqrt3}{2} =2-\sqrt3$