Дано: cos t= -5/13 pi/2 < t < pi Найти: sin t,tg t и ctg t

Решите задачу:

$sin \alpha = \sqrt{1-cos^2 \alpha } \\ \\ sint= \sqrt{1-\frac{25}{169} } = \sqrt{ \frac{169-25}{169} } = \sqrt{ \frac{144}{169} }=\frac{12}{13} \\ \\ tg \alpha = \frac{sin \alpha }{cos \alpha } \\ \\ tgt= \frac{\frac{12}{13}}{\frac{-5}{13} } =\frac{12}{13}*\frac{13}{-5}=\frac{12}{-5}=-2,4 \\ \\ ctg \alpha = \frac{cos \alpha }{sin \alpha } \\ \\ ctgt=\frac{\frac{-5}{13}}{\frac{12}{13}}=\frac{-5}{13}*\frac{13}{12}=\frac{-5}{12}=-\frac{5}{12}$