Помогите решить 5 номер

$cos \alpha =- \frac{9}{41}, \alpha \in( \frac{ \pi }{2}; \pi ) \Rightarrow a \in$ II четверти, значит:
$sin \alpha \ \textgreater \ 0 , tg \alpha \ \textless \ 0 , ctg \alpha \ \textless \ 0$
$sin \alpha = \sqrt{1-cos^2 \alpha }= \sqrt{1-(- \frac{9}{41})^2} = \sqrt{1- \frac{81}{1681} }= \sqrt{ \frac{1600}{1681} }= \frac{40}{41} \\ tg \alpha= \frac{sin \alpha }{cos \alpha } = \frac{\frac{40}{41} }{- \frac{9}{41}} =\frac{40}{41} } \cdot (- \frac{41}{9} )=- \frac{40}{9} =-4 \frac{4}{9} \\ ctg \alpha = \frac{1}{tg \alpha } = \frac{1}{- \frac{40}{9}} =- \frac{9}{40}$