Помогитеэто точно срочно,ребят

Решите задачу:

$\frac{2cos(\frac{\pi}{6}-2a)-\sqrt3\cdot sin(2,5\pi -2a)}{cos(4,5\pi -2a)+2cos(\frac{\pi}{6}+2a)} =\\\\=[\, sin(2,5\pi -2a)=sin(2\pi+\frac{\pi}{2}-2a)=sin(\frac{\pi}{2}-2a)=cos2a;\\\\cos(4,5\pi -2a)=cos(4\pi +\frac{\pi}{2}-2a)=cos(\frac{\pi}{2}-2a)=sin2a \, ]=$

$=\frac{2\cdot (cos(\frac{\pi}{6}-2a)-\frac{\sqrt3}{2}cos2a)}{2\cdot (\frac{1}{2}sin2a+cos(\frac{\pi}{6}+2a))} = \frac{cos\frac{\pi}{6}\cdot cos2a+sin\frac{\pi}{6}\cdot sin2a-\frac{\sqrt3}{2}cos2a}{\frac{1}{2}sin2a+cos\frac{\pi}{6}*cos2a-sin\frac{\pi}{6}*sin2a}=$

$= \frac{\frac{\sqrt3}{2}cos2a+\frac{1}{2}sin2a-\frac{\sqrt3}{2}cos2a}{\frac{1}{2}sin2a+\frac{\sqrt3}{2}cos2a-\frac{1}{2}sin2a} = \frac{\frac{1}{2}sin2a}{\frac{\sqrt3}{2}cos2a} = \frac{sin2a}{\sqrt3\cdot cos2a} = \frac{tg2a}{\sqrt3}$