Доказать тождество: 1) [latex]cosa*sin(6pi-a)*(1+ctg(-a)^2)=ctg(-a)[/latex] 2) [latex] \frac{1-sin(-a)^2}{cos(4pi-a)} * \frac{sin(a-2pi)}{1-cos(-a)^2} =ctga[/latex]

1) Sin(6π-α)=-Sinα; Ctg²(-α)=Ctg²α Получаем [latex] \frac{-Cos \alpha *Sin \alpha }{Sin^{2} \alpha}=-Ctg \alpha =Ctg(- \alpha ) [/latex] 2) Sin²(-α)=Sin²(α); Sin(α-2π)=Sinα; Cos(4π-α)=Cosα; Cos²(-α)=Cosα Получаем [latex] \frac{1-Sin \alpha^{2} }{Cos \alpha}* \frac{Sin \alpha }{1-Cos^{2} \alpha}= \frac{Cos^{2} \alpha*Sin \alpha}{Cos \alpha *Sin^{2} \alpha}=Ctg \alpha [/latex]