Упр.1291 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)
- Докажите, что $$\alpha-\beta=\frac{\pi}{4}$$, если $$\alpha$$ и $$\beta$$ — острые углы и $$\operatorname{tg}\alpha=2\frac{1}{3}$$ и $$\operatorname{tg}\beta=\frac{2}{5}$$.
- Найдите значение выражения:
а) $$\frac{\sqrt{(\cos 60^\circ-\sin 60^\circ)^2}}{\sin 30^\circ(1-\operatorname{tg}60^\circ)}$$;
б) $$\frac{\sqrt{\left(\operatorname{tg}\frac{\pi}{6}-\operatorname{tg}\frac{\pi}{3}\right)^2}}{\sqrt{\left(\operatorname{ctg}\frac{\pi}{6}-\operatorname{ctg}\frac{\pi}{3}\right)^2}}$$.
а)
$$\frac{\sqrt{(\cos 60^\circ-\sin 60^\circ)^2}}{\sin 30^\circ\,(1-\tg 60^\circ)} = \frac{\sqrt{\left(\frac12-\frac{\sqrt3}{2}\right)^2}}{\frac12\,(1-\sqrt3)}$$
$$\sqrt{\left(\frac12-\frac{\sqrt3}{2}\right)^2} = \left|\frac12-\frac{\sqrt3}{2}\right| = \frac{\sqrt3}{2}-\frac12$$
$$\frac{\frac{\sqrt3}{2}-\frac12}{\frac12\,(1-\sqrt3)} = \frac{\sqrt3-1}{1-\sqrt3} = -1$$
б)
$$\frac{\sqrt{\left(\tg \frac{\pi}{6}-\tg \frac{\pi}{3}\right)^2}}{\sqrt{\left(\ctg \frac{\pi}{6}-\ctg \frac{\pi}{3}\right)^2}} = \frac{\sqrt{\left(\frac{\sqrt3}{3}-\sqrt3\right)^2}}{\sqrt{\left(\sqrt3-\frac{\sqrt3}{3}\right)^2}}$$
$$\sqrt{\left(\frac{\sqrt3}{3}-\sqrt3\right)^2} = \left|\frac{\sqrt3}{3}-\sqrt3\right| = \sqrt3-\frac{\sqrt3}{3}$$
$$\sqrt{\left(\sqrt3-\frac{\sqrt3}{3}\right)^2} = \left|\sqrt3-\frac{\sqrt3}{3}\right| = \sqrt3-\frac{\sqrt3}{3}$$
$$\frac{\sqrt3-\frac{\sqrt3}{3}}{\sqrt3-\frac{\sqrt3}{3}}=1$$
Ответ
а) $$-1$$; б) $$1$$.
