Question

срочно sin ^(2)2a-4 sin^(2)2a/sin²2a +4 sin²a-4​

Answer 1

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

Answer 2

Ответ:

$\dfrac{sin^22a-4sin^2a}{sin^22a+4sin^2a-4}=\dfrac{4sin^2a\cdot cos^2a-4sin^2a}{4sin^2a\cdot cos^2a-4(1-sin^2a)}=\dfrac{4sin^2a\cdot (cos^2a-1)}{4sin^2a\cdot cos^2a-4cos^2a}=\\\\\\=\dfrac{4sin^2a\cdot (-sin^2a)}{4cos^2a\cdot (sin^2a-1)}=\dfrac{-4\, sin^4a}{4cos^2a\cdot (-cos^2a)}=\dfrac{sin^4a}{cos^4a}=tg^4a$