Question

Помогите пожалуйста, все кроме 3 и 8

Answer 1

$1); frac{1+sin2a}{(sina+cosa)^2}=frac{1+sin2a}{sin^2a+cos^2a+2sina, cosa}=frac{1+sin2a}{1+sin2a}=1; ; ,; ; 1=1\\\2); frac{sin^3a-sinacdot cos^2a}{sin^4a-cos^4a}=frac{sinacdot (sin^2a-cos^2a)}{(sin^2a-cos^2a)(sin^2a+cos^2a)}=frac{sina}{1}=sina; ; ,; ; sina=sina$

$4); frac{sin^2a}{cosa(1+ctga)}-frac{cos^2a}{sina(1+tga)}=frac{sin^2acdot sina}{cosacdot (sina+cosa)}-frac{cos^2acdot cosa}{sinacdot (cosa+sina)}=\\=frac{sin^4a-cos^4a}{sinacdot cosacdot (sina+cosa)}=frac{(sina-cosa)(sina+cosa)(sin^2a+cos^2a)}{sinacdot cosacdot (sina+cosa)}=\\=frac{sina-cosa}{1/2cdot sin2a}=frac{sina-sin(frac{pi}{2}-a)}{1/2cdot sin2a}=frac{2cdot 2sin(a-frac{pi}{4})cdot cosfrac{pi}{4}}{sin2a}=frac{2sqrt2cdot sin(a-frac{pi}{4})}{sin2a}$

$5); 2cos^2(frac{pi}{4}+a)+sin2a=2cdot (frac{sqrt2}{2}cdot cosa-frac{sqrt2}{2}cdot sina)^2+sin2a=\\=2cdot frac{2}{4}cdot (cos^2a+sin^2a-2sinacdot cosa)+sin2a=\\=1-sin2a+sin2a=1; ; ,; ; 1=1\\\6); ; sin2x-tgx=2sinxcdot cosx-frac{sinx}{cosx}=frac{2sinxcdot cos^2x-sinx}{cosx}=\\=frac{sinx(2cos^2x-1)}{cosx}=frac{sinxcdot cos2x}{cosx}=tgxcdot cos2x\\\7); tg(frac{3pi}{2}+a)+2ctg(frac{pi}{2}-a)=-tga+2ctga=-frac{sina}{cosa}+frac{2cosa}{sina}=$

$=frac{2cos^2a-sin^2a}{sinacdot cosa}=frac{2(1-sin^2a)-sin^2a}{sinacdot cosa}=frac{2-3sin^2a}{sinacdot cosa}$