Question

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

Answer 1

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

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

Answer 2

1)sin²2a-4sin²a=4sin²acos²a-4sin²a=4sin²a(cos²a-1)=4sin²a*(-sin²a)=
=-4sin^4a
2)sin²2a-4+4sin^4a=sin²2a-4(1-sin^4)a=sin²2a-4(1-sin²a)(1+sin²a)=
=4sin²acos²a-4*cos²a*(1+sin²a)=4sin²acos²a-4cos²a-4cos²asin²a=-4cos²a
3)-4sin^4a/(-4cos²a)=tg²a*sin²a