Question

помогите тригонометрия очень надо без разницы 1 или 2​

Answer 1

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