Question

Вычислите: arctg (-√3) + arccos (-√3/2) + arcsin 1

Answer 1

$\arctan(-\sqrt{3})+\arccos\left(-\frac{\sqrt3}{2}\right)+\arcsin 1=-\frac{\pi}{3}+\frac{5\pi}{6}+\frac{\pi}{2}=$

$=-\frac{\pi}{3}+\frac{\pi}{2}+\frac{5\pi}{6}=\frac{\pi}{6}+\frac{5\pi}{6}=\pi$