Question

СРОЧНО !!! 50 баллов. sin^2 2x - 4 sin^2x / sin^2 2x+4 sin ^2 x-4 =tg ^4x

Answer 1

$\displaystyle\bf\\\frac{Sin^{2}2x-4Sin^{2} x }{Sin^{2}2x+4Sin^{2} x-4 } =\\\\\\=\frac{(2Sinx Cosx)^{2} -4Sin^{2} x }{(2Sinx Cosx)^{2} +4Sin^{2} x-4 \cdot(Sin^{2} x+Cos^{2}x) } =\\\\\\=\frac{4Sin^{2} x Cos^{2} x -4Sin^{2} x }{4Sin^{2} x Cos^{2} x+4Sin^{2} x-4 Sin^{2} x-4Cos^{2}x } =\\\\\\=\frac{4Sin^{2} x \cdot(Cos^{2} x -1)}{4Sin^{2} x Cos^{2} x-4Cos^{2}x } =\frac{4Sin^{2} x \cdot(-Sin ^{2} x )}{4Cos^{2} x\cdot(Sin^{2}x -1)} =$

$\displaystyle\bf\\=\frac{-4Sin^{4} x}{4Cos^{2} x\cdot(-Cos^{2} x)}=\frac{-Sin^{4}x }{-Cos^{4} x} =tg^{4} x\\\\\\tg^{4} x=tg^{4} x$

Тождество доказано