Question

sin^2a/tg^2a + cos^2a/ctg^2a- 1/sin^2a=-ctg^2a​

Answer 1

$\frac{Sin^{2}\alpha}{tg^{2}\alpha}+\frac{Cos^{2}\alpha}{Ctg^{2}\alpha}-\frac{1}{Sin^{2}\alpha}=\frac{Sin^{2}\alpha Cos^{2}\alpha}{Sin^{2}\alpha}+\frac{Cos^{2}\alpha Sin^{2}\alpha}{Cos^{2}\alpha}-\frac{1}{Sin^{2}\alpha} =Cos^{2}\alpha+Sin^{2}\alpha-\frac{1}{Sin^{2}\alpha}=1-\frac{1}{Sin^{2}\alpha} =\frac{Sin^{2}\alpha-1}{Sin^{2}\alpha}=\frac{-Cos^{2}\alpha}{Sin^{2}\alpha}=-Ctg^{2} \alpha\\\\-Ctg^{2}\alpha=-Ctg^{2}\alpha$

Тождество доказано