\( (1/6)^2 + \cos^2 a = 1 \)
\( 1/36 + \cos^2 a = 1 \)
\( \cos^2 a = 1 - 1/36 = 35/36 \)
Поскольку \( a \) принадлежит второй четверти (\( \pi/2 < a < \pi \)), \( \cos a \) отрицателен.
\( \cos a = -\sqrt{35/36} = -\frac{\sqrt{35}}{6} \)
\( \operatorname{tg} a = \frac{\sin a}{\cos a} = \frac{1/6}{-\sqrt{35}/6} = -\frac{1}{\sqrt{35}} = -\frac{\sqrt{35}}{35} \)
\( \operatorname{ctg} a = \frac{1}{\operatorname{tg} a} = \frac{1}{-1/\sqrt{35}} = -\sqrt{35} \)
Ответ: \( \cos a = -\frac{\sqrt{35}}{6} \), \( \operatorname{tg} a = -\frac{\sqrt{35}}{35} \), \( \operatorname{ctg} a = -\sqrt{35} \)