Question

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Answer 1

$arcsin(-\frac{1}{2})-arctg\sqrt3=-arcsin\frac{1}{2}-\frac{\pi }{3}=-\frac{\pi }{6}-\frac{\pi }{3}=-\frac{3\pi }{6}=-\frac{\pi }{2}\\\\\\P.S.\; \; sin\frac{\pi }{6}=\frac{1}{2}\; \; \to \; \; arcsin\frac{1}{2}=\frac{\pi }{6}\; ,\\\\tg\frac{\pi }{3}=\sqrt3\; \; \to \; \; arctg\sqrt3=\frac{\pi }{3}\; ,\\\\arcsin(-\alpha )=-arcsin\alpha \; .$

Answer 2

$arcSin(-\frac{1}{2})-arctg \sqrt{3} =-arcSin\frac{1}{2}-arctg \sqrt{3}=- \frac{\pi} {6}- \frac{\pi} {3}=- \frac{\pi} {6}- \frac{2\pi} {6}=- \frac{3\pi} {6} =-\frac{\pi} {2}$