Упр.1190 ГДЗ Колягин Ткачёва 10 класс (Алгебра)

1) arctg(ctg 5пи/6);
2) arctg(ctg 3пи/4);
3) arctg(2sin пи/3).

1. $$\operatorname{arctg}\left(\operatorname{ctg}\frac{5\pi}{6}\right) =\operatorname{arctg}\left(\operatorname{ctg}\left(\frac{\pi}{2}+\frac{\pi}{3}\right)\right)$$

   $$\operatorname{ctg}\left(\frac{\pi}{2}+\alpha\right)=-\operatorname{tg}\alpha,$$
   значит
   $$\operatorname{arctg}\left(\operatorname{ctg}\frac{5\pi}{6}\right) =\operatorname{arctg}\left(-\operatorname{tg}\frac{\pi}{3}\right) =-\operatorname{arctg}\left(\operatorname{tg}\frac{\pi}{3}\right) =-\frac{\pi}{3}.$$

2. $$\operatorname{arctg}\left(\operatorname{ctg}\frac{3\pi}{4}\right) =\operatorname{arctg}\left(\operatorname{ctg}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)\right)$$

   $$\operatorname{ctg}\left(\frac{\pi}{2}+\alpha\right)=-\operatorname{tg}\alpha,$$
   значит
   $$\operatorname{arctg}\left(\operatorname{ctg}\frac{3\pi}{4}\right) =\operatorname{arctg}\left(-\operatorname{tg}\frac{\pi}{4}\right) =-\operatorname{arctg}\left(\operatorname{tg}\frac{\pi}{4}\right) =-\frac{\pi}{4}.$$

3. $$\operatorname{arctg}\left(2\sin\frac{\pi}{3}\right) =\operatorname{arctg}\left(2\cdot\frac{\sqrt{3}}{2}\right) =\operatorname{arctg}\sqrt{3} =\frac{\pi}{3}.$$

Ответ

$$-\frac{\pi}{3};\ -\frac{\pi}{4};\ \frac{\pi}{3}.$$