Вычислите: 1) arctg √3 - arctg1 + arcctg (-√3) 2) tg (arcctg √3/3) 3) arcctg (sin π/2)

Решите задачу:

$\arctan(\sqrt3)-\arctan (1)+arcctg(-\sqrt3)= \frac{\pi}{3}- \frac{ \pi }{4}+ \frac{5 \pi }{6}= \frac{4 \pi-3\pi+10\pi}{12}=\\\\= \frac{11 \pi }{12} \\\\\\ \tan (\arctan( \frac{\sqrt3}{3})=\tan( \frac{\pi}{6})= \frac{\sqrt 3}{3}\\\\\\ arcctg(\sin \frac{\pi}{2})=arcctg(1)= \frac{\pi}{4}$