Алгебра 10 класс) Помогите с № 30

Решите задачу:

$1)\quad \frac{tg\frac{\pi}{8}}{1-tg^2\frac{\pi}{8}}=[\; tg2a= \frac{2tga}{1-tg^2a} \; \frac{tga}{1-tg^2a}=\frac{1}{2}tg2a]=\frac{1}{2}tg\frac{\pi}{4}=\frac{1}{2}\cdot 1=\frac{1}{2}\\\\2)\quad \frac{1-tg^2\frac{\pi}{12}}{1+tg^2\frac{\pi}{12}} =\left [cosa= \frac{1-tg^2\frac{a}{2}}{1+tg^2\frac{a}{2}}\right ]= cos\frac{\pi}{6}=\frac{\sqrt3}{2}$

$3)\quad \frac{tg\frac{\pi}{8}}{1+tg^2\frac{\pi}{8}} =\frac{1}{2}sin(2\cdot \frac{\pi}{8})=\frac{1}{2}\cdot sin\frac{\pi}{4}=\frac{1}{2}\cdot \frac{\sqrt2}{2}=\frac{\sqrt2}{4}$

$4)\quad \frac{sin25\cdot sin65}{cos40} = \frac{sin25\cdot sin(90-25)}{cos40} = \frac{sin25\cdot cos25}{cos40} = \frac{\frac{1}{2}sin50}{cos(90-50)}=\\\\= \frac{sin50}{2sin50} = \frac{1}{2}\\\\5)\quad (cos^210-cos ^280)^2+cos^270=\\\\=(cos^210-cos^2(90-10))^2+cos^270=\\\\=(cos^210-sin^210)^2+cos^270=cos^220+cos^2(90-20)=\\\\=cos^220+sin^220=1$