Question

помогите пожалуйста по алгебре срочно обведенные под в и г​

Answer 1

$21.31)arctg\sqrt{3}=\frac{\pi }{3}\\\\arctg(-\frac{1}{\sqrt{3}})=-\frac{\pi }{6}\\\\\\21.32)arcCtg(-\frac{\sqrt{3}}{3})=\frac{2\pi }{3}\\\\arcCtg0=\frac{\pi }{2}$

$21.33)arcCtg(-\frac{\sqrt{3}}{3})-arctg\frac{\sqrt{3}}{3}=\frac{2\pi }{3}-\frac{\pi}{6}=\frac{4\pi -\pi}{6}=\frac{3\pi}{6}=\frac{\pi }{2} \\\\arcCos(-\frac{1}{2})-arctg(-\sqrt{3})=\frac{2\pi }{3}-(-\frac{\pi }{3})=\frac{2\pi }{3}+\frac{\pi }{3}=\frac{3\pi }{3}=\pi$

$21.34)arctg(-\sqrt{3} )+arcCos(-\frac{\sqrt{3}}{2})+arcSin1=-\frac{\pi }{3}+\frac{5\pi }{6}+\frac{\pi }{2}=\frac{-2\pi+5\pi+\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}\\\\arcSin(-1)-\frac{3}{2}arcCos\frac{1}{2}+3arcCtg(-\frac{\sqrt{3}}{3})=-\frac{\pi }{2}-\frac{3}{2}*\frac{\pi }{3}+3*\frac{2\pi }{3}=-\frac{\pi }{2}-\frac{\pi }{2}+2\pi=-\pi+2\pi=\pi$

$21.35)Cos(arctg0)=Cos0=1\\\\Ctg(arctg(-1))=Ctg(-\frac{\pi }{4})=-1$