127. B) arccos(-\frac{\sqrt{3}}{2})+arcsin(-\frac{\sqrt{3}}{2});

B) arccos($$-\frac{\sqrt{3}}{2}$$) + arcsin($$-\frac{\sqrt{3}}{2}$$) = $$\frac{5\pi}{6} + (-\frac{\pi}{3})$$ = $$\frac{5\pi - 2\pi}{6}$$ = $$\frac{3\pi}{6}$$ = $$\frac{\pi}{2}$$.

Ответ: $$\frac{\pi}{2}$$