Question

arccos(-1/2)+arcsin√2/2-arctg(-1)​

Answer 1

$arcCos(-\frac{1}{2})+arcSin\frac{\sqrt{2}}{2}-arctg(-1)=\frac{2\pi }{3}+\frac{\pi }{4}-(-\frac{\pi }{4}) =\frac{2\pi }{3}+\frac{\pi }{4}+\frac{\pi }{4} =\frac{2\pi }{3}+\frac{\pi }{2}=\frac{4\pi+3\pi}{6}=\frac{7\pi}{6}$