Sin^2 2a-4sin^2a/sin^2 2a+4sin^2a-4=tg^4 a

Решите задачу:

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