Ну пожалуйста! в и г

[latex]arcctg(-\frac{\sqrt3}{3})-arctg\frac{\sqrt3}{3}=\pi-arcctg\frac{\sqrt3}{3}-\frac{\pi}{6}=\pi-\frac{\pi}{3}-\frac{\pi}{6}=\\ =\frac{6\pi-2\pi-\pi}{6}= \frac{\pi}{2};\\\\arccos(-\frac{1}{2})-atcctg(-\sqrt{3})=\pi-arccos\frac{1}{2}-(\pi-arcctg\sqrt{3})=\\=\pi-\frac{\pi}{3}-\pi+\frac{\pi}{6}=-\frac{\pi}{6}.[/latex]