Question

((sina-cosa)^2-1+sin4a)/(cos2a+cos4a)

Answer 1

Для удобства написания заменила угол а на угол х


$\Large \frac{(sinx-cosx)^2-1+sin4x}{cos2x+cos4x} = \frac{sin^2x-2sinx*cosx+cos^2x-1+sin4x}{cos2x+cos4x}=\\= \frac{-2sinx*cosx+sin4x}{cos2x+(2cos^2x-1)}= \frac{sin4x-sin2x}{(cos2x+1)(2cos2x-1)}=\\= \frac{2sin2x*cos2x-sin2x}{(cos2x+1)(2cos2x-1)}= \frac{sin2x(2cos2x-1)}{(cos2x+1)(2cos2x-1)}= \frac{sin2x}{cos2x+1}=\\= \frac{2sinx*cosx}{cos^2-sin^2x+cos^2x+sin^2x}= \frac{2sinx*cosx}{2cos^2x}= \frac{sinx}{cosx}=tgx$

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пояснения
$\Large cos2x+2cos^22x-1$
разложим на множители

$\Large cos2x=t\\2t^2+t-1=0\\t_1=-1; t_2=1/2\\2cos^2x+cos2x-1=(cos2x+1)(2cos2x-1)$