Odpowiedź:
[tex](cos120 - ctg 150)^2= (cos(180-60) - ctg(180-30)^2=(-cos60 + ctg30)^2=(-\frac{1}{2}+\sqrt{3})^2=3-\sqrt{3} +\frac{1}{4} \\cos300:sin315=cos(360-60):sin(360-45)=cos60:-sin45=\frac{1}{2}:\frac{\sqrt{2}}{2}=\frac{\sqrt{2} }{2}\\(sin135+tg120)^2=(sin(180-45)+tg(180-60))^2=(sin45-tg60)^2=(\frac{\sqrt{2} }{2}-\sqrt{3})^2= \frac{1}{2}-\sqrt{6}+3[/tex]
