Question

помогите пожалуйста даю 30 баллов ​

Answer 1

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

$-tg^{6}\alpha=-tg^{6}\alpha$

Тождество доказано

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

- tgα = - tgα

Тождество доказано

$3)\frac{1+Cos(\pi+\alpha)}{-Sin(\alpha-\pi)}=\frac{1-Cos\alpha}{Sin\alpha}=\frac{2Sin^{2}\frac{\alpha }{2}}{2Sin\frac{\alpha }{2}Cos\frac{\alpha }{2}}=\frac{Sin\frac{\alpha }{2}}{Cos\frac{\alpha }{2}}=tg\frac{\alpha }{2}\\\\tg\frac{\alpha }{2}=tg\frac{\alpha }{2}$

Тождество доказано