А)b)c)d)

Решите задачу:

$arcctg1-arctg\sqrt3-arccos(-\frac{1}{2})=\frac{\pi}{4}-\frac{\pi}{3}-(\pi -\frac{\pi}{3})=-\frac{3\pi}{4}\\\\b)\; arcsin(-\frac{1}{2})+arctg(-\frac{1}{\sqrt3})-arcctg\sqrt3=-\frac{\pi}{6}-\frac{\pi}{6}-\frac{\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}\\\\c)\; arcsin(-1)-\frac{3}{2}arccos\frac{1}{2}+3arcctg(-\frac{1}{\sqrt3})=-\frac{\pi}{2}-\frac{3}{2}\cdot \frac{\pi}{3}+3(\pi -\frac{\pi}{3})=\\\\=-\frac{\pi}{2}-\frac{\pi}{2}+2\pi =-\pi +2\pi =\pi$

$d)\; -4arcsin\frac{\sqrt2}{2}+8arccos\frac{\sqrt2}{2}-15arctg\frac{\sqrt3}{3}=-4\cdot \frac{\pi}{4}+8\cdot \frac{\pi}{4}-15\cdot \frac{\pi}{6}=\\\\=-\pi +2\pi -\frac{15}{6}\pi =-\frac{9}{6}\pi =-\frac{3}{2}\pi$