Question

Помогите с уравнением

Answer 1

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

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).