Упр.1291 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) v(cos(60°)-sin(60°))^2/(sin(30°)(1-tg(60°));
б) v(tg(?/6)-tg(?/3))^2/v(ctg(?/6)-ctg(?/3))^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) $$1$$.