Question

Нужна помощь!
Алгебра 10 класс
Задание на фото
Даю 25 балов!
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Answer 1

$1)Sin^{2}2\alpha+Cos^{2}2\alpha+Ctg^{2}5\alpha=1+Ctg^{2}5\alpha=\frac{1}{Sin^{2}5\alpha} \\\\\\2)Sin\frac{\alpha }{3}*Ctg\frac{\alpha }{3}=Sin\frac{\alpha }{3}*\frac{Cos\frac{\alpha}{3} }{Sin\frac{\alpha}{3}}=Cos\frac{\alpha}{3}\\\\\\3)1-\frac{1}{Sin^{2}\alpha}=1-(1+Ctg^{2}\alpha)=1-1-Ctg^{2}\alpha=-Ctg^{2} \alpha$

$4)\frac{Sin^{2}\alpha-1}{Cos^{2}\alpha-1}+tg\alpha Ctg\alpha=\frac{-Cos^{2}\alpha}{-Sin^{2}\alpha}+1=Ctg^{2}\alpha+1=\frac{1}{Sin^{2}\alpha}\\\\\\5)(tg\alpha Cos\alpha)^{2}+(Ctg\alpha Sin\alpha)^{2}=(\frac{Sin\alpha}{Cos\alpha}*Cos\alpha)^{2}+(\frac{Cos\alpha}{Sin\alpha}*Sin\alpha)^{2}=Sin^{2}\alpha+Cos^{2}\alpha=1$

$6)\frac{Sin^{2}\alpha}{1+Ctg^{2}\alpha(Cos^{2}\alpha-1)}=\frac{Sin^{2}\alpha}{1+Ctg^{2}\alpha*(-Sin^{2}\alpha)}=\frac{Sin^{2}\alpha}{1-Cos^{2}\alpha}=\frac{Sin^{2}\alpha}{Sin^{2}\alpha}=1\\\\\\7)(\frac{1}{Cos\alpha}+tg\alpha)(\frac{1}{Cos\alpha}-tg\alpha)=(\frac{1}{Cos\alpha})^{2} -tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}-tg^{2}\alpha=1+tg^{2}\alpha-tg^{2}\alpha=1$

$8)(tg\beta +Ctg\beta )^{2} -(tg\beta-Ctg\beta)^{2}=tg^{2}\beta+2tg\beta Ctg\beta+Ctg^{2}\beta-tg^{2}\beta+2tg\beta Ctg\beta-Ctg^{2} \beta=4tg\beta Ctg\beta=4$