Question

637. Докажите тождество
Пожалуйста помогите
30-баллов!!!!!!​

Answer 1

$1)Cos^{4}\alpha -Sin^{4}\alpha=(Cos^{2}\alpha)^{2}-(Sin^{2}\alpha)^{2}=(Cos^{2}\alpha-Sin^{2}\alpha)\underbrace{(Cos^{2}\alpha+Sin^{2}\alpha)}_{1}=\\\\=\boxed{Cos^{2}\alpha-Sin^{2}\alpha}\\\\\\2)(Cos\alpha*tg\alpha)^{2} +(Sin\alpha*Ctg\alpha)^{2}=(Cos\alpha*\frac{Sin\alpha }{Cos\alpha})^{2} +(Sin\alpha*\frac{Cos\alpha }{Sin\alpha})^{2}=\\\\=Sin^{2}\alpha +Cos^{2}\alpha=\boxed1$

$3)Sin^{4}\alpha +Sin^{2} \alpha*Cos^{2}\alpha=Sin^{2}\alpha\underbrace{(Sin^{2}\alpha+Cos^{2} \alpha)}_{1}=Sin^{2}\alpha=\boxed{1-Cos^{2}\alpha}\\\\\\4)Ctg^{2}\alpha-tg^{2}\alpha=\frac{Cos^{2}\alpha}{Sin^{2}\alpha}-\frac{Sin^{2}\alpha }{Cos^{2}\alpha}=\frac{Cos^{4}\alpha-Sin^{4}\alpha }{Sin^{2}\alpha Cos^{2}\alpha}=\frac{(Cos^{2}\alpha-Sin^{2}\alpha)( Cos^{2}\alpha+Sin^{2}\alpha) }{Sin^{2}\alpha Cos^{2}\alpha}=$

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