Помогите пожалуйста, все кроме 3 и 8

Question 1

Question 2

6 и 7 ошибки ...

Question 3

Решите задачу:

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

$4)\; \frac{sin^2a}{cosa(1+ctga)}-\frac{cos^2a}{sina(1+tga)}=\frac{sin^2a\cdot sina}{cosa\cdot (sina+cosa)}-\frac{cos^2a\cdot cosa}{sina\cdot (cosa+sina)}=\\\\=\frac{sin^4a-cos^4a}{sina\cdot cosa\cdot (sina+cosa)}=\frac{(sina-cosa)(sina+cosa)(sin^2a+cos^2a)}{sina\cdot cosa\cdot (sina+cosa)}=\\\\=\frac{sina-cosa}{1/2\cdot sin2a}=\frac{sina-sin(\frac{\pi}{2}-a)}{1/2\cdot sin2a}=\frac{2\cdot 2sin(a-\frac{\pi}{4})\cdot cos\frac{\pi}{4}}{sin2a}=\frac{2\sqrt2\cdot sin(a-\frac{\pi}{4})}{sin2a}$

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

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