Question

Задание по алгебре, срочно (во вложении картинка)

Answer 1

$(Ctg^{2}\alpha-tg^{2}\alpha)*\frac{1}{Ctg^{2}\alpha-1}=(Ctg^{2}\alpha-\frac{1}{Ctg^{2}\alpha } )*\frac{1}{Ctg^{2}\alpha-1}=\\\\=\frac{Ctg^{4}\alpha-1}{Ctg^{2}\alpha}*\frac{1}{Ctg^{2}\alpha-1}=\frac{(Ctg^{2}\alpha-1)(Ctg^{2}\alpha+1)}{Ctg^{2}\alpha(Ctg^{2}\alpha -1)}=\frac{Ctg^{2}\alpha+1 }{Ctg^{2}\alpha}=\frac{1}{Sin^{2}\alpha}:\frac{Cos^{2}\alpha}{Sin^{2}\alpha}=\\\\=\frac{Sin^{2}\alpha}{Sin^{2}\alpha*Cos^{2}\alpha} =\frac{1}{Cos^{2}\alpha}$

$\boxed{\frac{1}{Cos^{2}\alpha} =\frac{1}{Cos^{2}\alpha}}$

Что и требовалось доказать