ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 77

\[y = \frac{1}{\left| \text{tg\ x} \right|};\ \ \ x = a.\]

\[1)\ a = \frac{5\pi}{6}:\]

\[\frac{1}{\left| \text{tg}\frac{5\pi}{6} \right|} = \frac{1}{\left| - tg\frac{\pi}{6} \right|} =\]

\[= 1\ :\left| - \frac{1}{\sqrt{3}} \right| = \sqrt{3};\]

\[2)\ a = \frac{3\pi}{4}:\]

\[\frac{1}{\left| \text{tg}\frac{3\pi}{4} \right|} = \frac{1}{\left| - tg\frac{\pi}{4} \right|} = 1\ :| - 1| = 1;\]

\[3)\ a = - \frac{\pi}{6}:\]

\[\frac{1}{\left| \text{tg}\left( - \frac{\pi}{6} \right) \right|} = \frac{1}{\left| - tg\frac{\pi}{6} \right|} =\]

\[= 1\ :\left| - \frac{1}{\sqrt{3}} \right| = \sqrt{3};\]

\[4)\ a = - \frac{2\pi}{3}:\]

\[\frac{1}{\left| \text{tg}\left( - \frac{2\pi}{3} \right) \right|} = \frac{1}{\left| \text{tg}\frac{\pi}{3} \right|} =\]

\[= 1\ :\left| \sqrt{3} \right| = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.\]