Question

Помогите, нужно доказать тождество:

Answer 1

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