Question

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

Answer 1

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x=1-3sin^2xcos^2x所以分子=3sin^2xcos^2x所以原式=3sin^2xcos^2x/sin^2x+3sin^2x=3cos^2x+3sin^2x=3(sin^2x...