Question

Помогите с уравнением

Answer 1

$\displaystyle \tt \frac{2cos(\alpha)^2\cdot tg\:\alpha}{sin(\alpha)^2-cos(\alpha)^2}=\frac{2cos(\alpha)^2\cdot\frac{sin\:\alpha}{cos\:\alpha}}{-cos(2\alpha)}=\frac{2cos(\alpha)sin(\alpha)}{-cos(2\alpha)}=\frac{sin(2\alpha)}{-cos(2\alpha)}=\bold{-tg(2\alpha)}$

Answer 2

Ответ:

Объяснение:

2*cos²α*tgα/(sin²α-cos²α)=2*cos²α*(sinα/cosα)/(-(cos²α-sin²α))=

=2*sinα*cosα/(-cos(2α))=-sin(2α)/cos(2α)=-tg(2α).