Помогите, нужно доказать тождество:

Решите задачу:

$\frac{(2sin \alpha *cos \alpha )^{2}-4sin ^{2} \alpha }{(2sin \alpha *cos \alpha )^{2}+4sin^{2} \alpha -4sin ^{2} \alpha -4cos ^{2} \alpha } =$
$\frac{4sin^{2} \alpha *cos^{2} \alpha-4sin ^{2} \alpha }{4sin^{2} \alpha *cos^{2} \alpha -4cos ^{2} \alpha }=$
$\frac{4sin^{2} \alpha *(cos^{2} \alpha-1 ) }{4*cos^{2} \alpha(sin^{2} \alpha -1) }= \frac{4sin^{2} \alpha *(cos^{2} \alpha-cos ^{2} \alpha -sin ^{2} \alpha ) }{4*cos^{2} \alpha(sin^{2} \alpha -sin ^{2} \alpha -cos ^{2} \alpha ) }=$
$\frac{sin^{2} \alpha *( -sin ^{2} \alpha ) }{cos^{2} \alpha( -cos ^{2} \alpha ) }= tg^{2} \alpha *tg ^{2} \alpha =tg ^{4} \alpha$