Question

Помогите решить
((cos^4(2A)-sin^4(2A))/cos(4A))-(cos(2A)-sin(2A))^2

Answer 1

$\frac{cos^42x-sin^42x}{cos4x}-(cos2x-sin2x)^2= \frac{(cos^22x-sin^22x)(cos^22x+sin^22x)}{cos^22x-sin^22x}- \\ -cos^22x+2sin2xcos2x-sin^22x=cos^22x+sin^22x-cos^22x+ \\ +2sin2xcos2x-sin^22x=2sin2xcos2x=sin4x$