Sin²2a-4sin²a/sin²2a-4+4sin ( чевёртой степени) a

Решите задачу:

$\dfrac{sin^22 \alpha - 4sin^2 \alpha }{sin^22 \alpha - 4 + 4sin^4 \alpha } = \dfrac{(sin2 \alpha )^2 - 4sin^2 \alpha }{sin^22 \alpha - 4(1 - sin^4 \alpha )} = \\ \\ \\ = \dfrac{4sin^2 \alpha cos^2 \alpha - 4sin^2 \alpha }{sin^22 \alpha - 4(1 - sin^2 \alpha )(1 + sin^2 \alpha )} = \\ \\ \\ = \dfrac{4sin^2 \alpha cos^2 \alpha - 4sin^2 \alpha } {4sin^2 \alpha cos^2 \alpha - 4cos^2 \alpha (1 + sin^2 \alpha) } =$

$= \dfrac{4sin^2 \alpha (cos^2 \alpha -1) }{4sin^2 \alpha cos^2 \alpha - 4sin^2 \alpha cos^2 \alpha - 4cos^2 \alpha } = \\ \\ \\ \dfrac{-4sin^4 \alpha }{-4cos^2 \alpha } = sin^2 \alpha \cdot tg^2 \alpha$

Sin²﻿2a-4sin²a/sin²2a-4+4sin ( чевёртой степени) a

Sin²2a-4sin²a/sin²2a-4+4sin ( чевёртой степени) a