Question

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+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+cos^2x)

=3