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

$2)(tgalpha+Ctgalpha)(1+Cosalpha)(Cosalpha-1)=(frac{Sinalpha }{Cosalpha}+frac{Cosalpha }{Sinalpha})*(Cos^{2}alpha-1)=frac{Sin^{2} alpha+Cos^{2}alpha}{Sinalpha Cosalpha}*(-Sin^{2}alpha)=-frac{1}{Sinalpha Cosalpha}*(Sin^{2}alpha)=-frac{Sinalpha}{Cosalpha }=-tgalpha$

- tgα = - tgα

Тождество доказано

$3)frac{1+Cos(pi+alpha)}{-Sin(alpha-pi)}=frac{1-Cosalpha}{Sinalpha}=frac{2Sin^{2}frac{alpha }{2}}{2Sinfrac{alpha }{2}Cosfrac{alpha }{2}}=frac{Sinfrac{alpha }{2}}{Cosfrac{alpha }{2}}=tgfrac{alpha }{2}\\tgfrac{alpha }{2}=tgfrac{alpha }{2}$

Тождество доказано