Алгебра 10 класс плизз

$\frac{2ctg2a}{tga+ctga}=cos2a\\\\ \frac{2ctg2a}{ \frac{sina}{cosa} + \frac{cosa}{sina} }=cos2a\\\\ \frac{2ctg2a}{ \frac{sin^2a+cos^2a}{sinacosa} } =cos2a\\\\ \frac{2ctg2a}{ \frac{1}{sinacosa} }=cos2a\\\\ctg2a*(2sinacosa)=2cosa\\\\ \frac{cos2a}{sin2a}*sin2a=cos2a\\\\cos2a=cos2a$
Тождество доказано

$\frac{2cos( \frac{3 \pi }{2}+a)-sin2a }{cos^2a-sin( \frac{ \pi }{2}-a) }-2tga= \frac{2sina-sin2a}{cos^2a-cosa}-2tga=\\\\= \frac{2sina-2sinacosa}{cosa(cosa-1)}-2tga= \frac{2sina(1-cosa)}{cosa(cosa-1)}-2tga=\\\\=- \frac{2sina}{cosa}-2tga=-2tga-2tga=-4tga$

$\frac{sin4x+(sin2x-cos2x)^2}{tg2x-ctg2x}= \frac{sin4x+sin^22x+cos^22x-2sin2xcos2x}{ \frac{sin2x}{cos2x}- \frac{cos2x}{sin2x} }=\\\\ \frac{sin4x+1-sin4x}{ \frac{sin^22x-cos^22x}{sin2xcos2x} }= \frac{1}{ \frac{-cos4x}{0,5sin4x} }=-0,5tg4x\\\\x= \frac{3 \pi }{16}\\\\-0,5tg(4* \frac{3 \pi }{16})=-0,5tg \frac{3 \pi }{4}=-0,5*(-1)=0,5$