Sin^2(-30градусов)-2ctg(-30градусов)-1/ /2-tg45градусов+4cos^2(-60градусов)

Решите задачу:

$\frac{sin^2(-30)-2ctg(-30)-1}{2-tg45+4cos^2(-60)} = \frac{(\frac{1}{2})^2-2\cdot (-\sqrt3)-1}{2-1+4\cdot (\frac{1}{2})^2} =\\\\= \frac{\frac{1}{4}+2\sqrt3-1}{2-1+1}= \frac{2\sqrt3-\frac{3}{4}}{2} = \frac{8\sqrt3-3}{4\cdot 2} = \frac{8\sqrt3-3}{8} =\sqrt3-\frac{3}{8}$