Question

HELP! ПОЖАЛУЙСТА Решите за 12 часов

Answer 1

$1)quad frac{sina-sin^3a}{cos^3a}=frac{sina(1-sin^2a)}{cos^3a}=frac{sinacdot cos^2a}{cos^3a}=frac{sina}{cosa}=tga\\2)quad frac{1}{tga}+frac{sina}{1+cosa}=frac{cosa}{sina}+frac{2sinfrac{a}{2}cdot cosfrac{a}{2}}{2cos^2frac{a}{2}}=ctga+frac{sinfrac{a}{2}}{cosfrac{a}{2}}=ctga+tgfrac{a}{2};\\ili\\frac{1}{tga}+frac{sina}{1+cosa}=frac{1+cosa+sinacdot tga}{tga(1+cosa)}= frac{cosa+cos^2a+sin^2a}{cosacdot tga(1+cosa)} = \\=frac{cosa+1}{sina(1+cosa)}=frac{1}{sina} ;$

$3)quad (1-sina-cosa)(1+sina+cosa)=\\=(1-(sina+cosa))(1+(sina+cosa))=\\=1^2-(sina+cosa)^2=1-(sin^2a+cos^2a+2sinacdot cosa)=\\=1-(1+sin2a)=-sin2a\\4)quad frac{1-4sin^2a, cos^2a}{(sina-cosa)^2}-2sina, cosa=\\= frac{1-(2sina, cosa)^2}{sin^2a+cos^2a-2sinacdot cosa}-sin2a=\\= frac{1-(sin2a)^2-(1-sin2a)sin2a}{1-sin2a}=frac{1-sin^22a-sin2a+sin^22a}{1-sin2a}=1$

$5)quad (2sin^2a+cos^2a-1)frac{1}{tg^2a}=(2sin^2a-underbrace{(1-cos^2a)}_{sin^2a})cdot frac{1}{tg^2a}=\\=sin^2acdot frac{cos^2a}{sin^2a}=cos^2a$