Question

Ну пожалуйста!
в и г

Answer 1

$arcctg(-frac{sqrt3}{3})-arctgfrac{sqrt3}{3}=pi-arcctgfrac{sqrt3}{3}-frac{pi}{6}=pi-frac{pi}{3}-frac{pi}{6}=\ =frac{6pi-2pi-pi}{6}= frac{pi}{2};\\arccos(-frac{1}{2})-atcctg(-sqrt{3})=pi-arccosfrac{1}{2}-(pi-arcctgsqrt{3})=\=pi-frac{pi}{3}-pi+frac{pi}{6}=-frac{pi}{6}.$