Question

Только правильно! помогите пожалуйста 2)​

Answer 1

ctg (x) = cos(x) / sin(x)

tg (x) = sin(x) / cos(x)

tg(x)*(cos(x) + ctg(x))*(1 - sin(x)) = sin(x)/cos(x) * (cos(x) + cos(x)/sin(x))*(1 - sin(x)) =

sin(x)/cos(x) *cos(x) (1 + 1/sin(x))*(1 - sin(x)) =

sin(x)*(sin(x) + 1)/sin(x)*(1 - sin(x)) =

(1 + sin(x))*(1 - sin(x)) = 1 - sin²(x) = cos²(x)

Answer 2

$\displaystyle\bf\\1)\\\\(Sinx-Cosx)(Sinx+Cosx)=Sin^{2} x-Cos^{2} x=\\\\=-(Cos^{2} x-Sin^{2} x)=-Cos2x\\\\\\2)\\\\tgx(Cosx+Ctgx)(1-Sinx)=(tgx\cdot Cosx+tgx\cdot Ctgx)(1-Sinx)=\\\\=\Big(\frac{Sinx}{Cosx} \cdot Cosx+tgx\cdot Ctgx\Big)\Big(1-Sinx\Big)=(Sinx+1)(1-Sinx)=\\\\=1^{2} -Sin^{2} x=1-Sin^{2} x=Cos^{2} x\\\\\\3)\\\\(Sin\alpha -Cos\alpha )^{2} +Sin2\alpha =Sin^{2} \alpha -2Sin\alpha Cos\alpha +Cos^{2} \alpha +Sin2\alpha =\\\\=1-Sin2\alpha +Sin2\alpha=1$

$\displaystyle\bf\\4)\\\\(1+tg\alpha )^{2} -sec^{2}\alpha =1+2tg\alpha +tg^{2}\alpha -\frac{1}{Cos^{2} \alpha } =\\\\=\frac{1}{Cos^{2} \alpha } +2tg\alpha -\frac{1}{Cos^{2} \alpha } =2tg\alpha$