Question

arccos(-1/2)+arcsin√2/2-arctg(-1)​

Answer 1

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