1. Вычислите: a) sin + 2cos-tg 45°-ctg 3 π π 6 π 2 = 6) cos(-)+tg(-)- ctg(-)=3 π π π 4

a) $$sin \frac{\pi}{3} + 2cos \frac{\pi}{6} - tg 45^\circ - ctg \frac{\pi}{2} = \frac{\sqrt{3}}{2} + 2 \cdot \frac{\sqrt{3}}{2} - 1 - 0 = \frac{\sqrt{3}}{2} + \sqrt{3} - 1 = \frac{3\sqrt{3}}{2} - 1$$

б) $$cos(-\frac{\pi}{3}) + tg(-\frac{\pi}{6}) - ctg(-\frac{\pi}{4}) = cos(\frac{\pi}{3}) - tg(\frac{\pi}{6}) + ctg(\frac{\pi}{4}) = \frac{1}{2} - \frac{\sqrt{3}}{3} + 1 = \frac{3 - 2\sqrt{3} + 6}{6} = \frac{9 - 2\sqrt{3}}{6}$$

Ответ: a) $$\frac{3\sqrt{3}}{2} - 1$$; б) $$\frac{9 - 2\sqrt{3}}{6}$$