\frac{sin120^0+sin300^0}{tg(-225^0)}
źródło:
\frac{sin120^0+sin300^0}{tg(-225^0)}=
=\frac{sin(90^0+30^0)+sin(360^0-60^0)}{-tg(180^0+45^0)}=
=\frac{cos30^0-sin60^0}{-tg45^0}=
=\frac{\frac{\sqrt3}{2}-\frac{\sqrt3}{2}}{-1}=0