化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
问题描述：
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
答
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