Question

Выражение ((sina-cosa)^2 - 1) (1+tg^2a) + 2ctg( п/2 - a)

Answer 1

$((sin \alpha -cos \alpha )^2-1)(1+tg^2 \alpha )+2ctg( \frac{ \pi }{2}- \alpha )= \\ \\ =(sin^2 \alpha +cos^2 \alpha-2sin \alpha cos \alpha -1 )* \frac{1}{cos^2 \alpha } +2tg \alpha = \\ \\ =(1-2sin \alpha cos \alpha-1)* \frac{1}{cos^2 \alpha } +2tg \alpha =- \frac{2sin \alpha cos \alpha}{cos^2 \alpha } +2tg \alpha = \\ \\ =- \frac{2sin \alpha }{cos \alpha } +2tg \alpha =-2tg \alpha +2tg \alpha =0$