[tex]tg120=tg(180-60)=-tg60\\ tg150=tg(180-30)=-tg30\\ \\ \frac{\frac{tg120+tg150}{sin60}}{sin90}=\frac{\frac{-tg60-tg30}{sin60}}1=(-\sqrt3-\frac{\sqrt3}3):\frac{\sqrt3}2=(-\frac{3\sqrt3}3-\frac{\sqrt3}3)*\frac2{\sqrt3}=-\frac{4\sqrt3}3*\frac2{\sqrt3}=-\frac{8\sqrt3}{3\sqrt3}=-\frac83[/tex]