HELP! ПОЖАЛУЙСТА Решите за 12 часов

Решите задачу:

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

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

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