Доказать торжество cos4α×tg2α-sin4α=-tg2α

Решите задачу:

$cos4\alpha*tg2\alpha-sin4\alpha=-tg2\alpha$
$\frac{sin2\alpha}{cos2\alpha}*cos4\alpha-sin4\alpha=-\frac{sin2\alpha}{cos2\alpha}$
$\frac{cos4\alpha*sin2\alpha-cos2\alpha*sin4\alpha}{cos2\alpha}=-\frac{sin2\alpha}{cos2\alpha}$
$cos4\alpha*sin2\alpha-cos2\alpha*sin4\alpha=-sin2\alpha$
$(\frac{1}{2}(sin(4\alpha+2\alpha)-sin(4\alpha-2\alpha)))-(\frac{1}{2}(sin(2\alpha+4\alpha)-sin(2\alpha-4\alpha)))$
$=-sin2\alpha$
$(\frac{1}{2}(sin6\alpha-sin2\alpha))-(\frac{1}{2}(sin6\alpha-sin(-2\alpha)))=-sin2\alpha$
$\frac{sin6\alpha-sin2\alpha}{2}-\frac{sin(-2\alpha)-sin6\alpha}{2}=-sin2\alpha$
$\frac{sin6\alpha}{2}-\frac{sin2\alpha}{2}-\frac{sin2\alpha}{2}-\frac{sin6\alpha}{2}=-sin2\alpha$
$-sin2\alpha=-sin2\alpha$