化简 (1+sin2x-cos2x)/(1+sin2x+cos2x)

问题描述:

化简 (1+sin2x-cos2x)/(1+sin2x+cos2x)

(')表示2次方 原式=(-2cos2x)/(1+sin2x+cos2x)+1 =<(-2cos'x+2sin'x)/(1+2sinxcox+cos'x-sin'x)>十1 =<2(sinx+cosx)(sinx-cosx)>/<(cosx+sinx)(cosx+sinx+cosx-sinx) =2(sinx-cosx)/2cosx+1 =tanx-1+1 =tanx

(1+sin2x-cos2x)/(1+sin2x+cos2x)
=(2sin^2x +2sinxcosx)/(2cos^2x+2sinxcosx)
=2sinx(sinx+cosx)/2cosx(cosx+sinx)
=tanx