Question

помогите
это точно срочно,ребят

Answer 1

$frac{2cos(frac{pi}{6}-2a)-sqrt3cdot sin(2,5pi -2a)}{cos(4,5pi -2a)+2cos(frac{pi}{6}+2a)} =\\=[, sin(2,5pi -2a)=sin(2pi+frac{pi}{2}-2a)=sin(frac{pi}{2}-2a)=cos2a;\\cos(4,5pi -2a)=cos(4pi +frac{pi}{2}-2a)=cos(frac{pi}{2}-2a)=sin2a , ]=$

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

$= frac{frac{sqrt3}{2}cos2a+frac{1}{2}sin2a-frac{sqrt3}{2}cos2a}{frac{1}{2}sin2a+frac{sqrt3}{2}cos2a-frac{1}{2}sin2a} = frac{frac{1}{2}sin2a}{frac{sqrt3}{2}cos2a} = frac{sin2a}{sqrt3cdot cos2a} = frac{tg2a}{sqrt3}$