Помогите решить . Тригонометрия 9 класс

Решите задачу:

$\frac{Sin ^{2}2 \alpha -4Cos ^{2} \alpha }{Sin ^{2}2 \alpha +4Cos ^{2} \alpha -4 } = \frac{4Sin ^{2} \alpha Cos ^{2} \alpha -4Cos ^{2} \alpha }{4Sin ^{2} \alpha Cos ^{2} \alpha +4Cos ^{2} \alpha -4 }= \frac{4Cos ^{2} \alpha (Sin ^{2} \alpha -1) }{4Sin ^{2} \alpha Cos ^{2} \alpha +4(Cos ^{2} \alpha -1) } =$ $\frac{-4Cos ^{3} \alpha }{4Sin ^{2} \alpha Cos ^{2} \alpha -4Sin ^{2} \alpha }= \frac{-4Cos ^{3} \alpha }{4Sin ^{2} \alpha (Cos ^{2} \alpha -1) } = \frac{-4Cos ^{3} \alpha }{-4Sin ^{3} \alpha } = ctg ^{3} \alpha$