搜搜做题作业网怎么证明cos2x/1-sin2x = cosx+sinx/cosx-sinx
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怎么证明cos2x/1-sin2x = cosx+sinx/cosx-sinx
证：
(cosx+sinx)/(cosx-sinx)
=[(cosx+sinx)(cosx-sinx)]/(cosx-sinx)^2
=[(cosx)^2-(sinx)^2]/[(cosx)^2-2sinxcosx+(sinx)^2]
=cos(2x)/(1-2sinxcosx)
=cos(2x)/[1-sin(2x)]
(cosx+sinx)/(cosx-sinx)
=[(cosx+sinx)(cosx-sinx)]/(cosx-sinx)^2
=[(cosx)^2-(sinx)^2]/[(cosx)^2-2sinxcosx+(sinx)^2]
=cos(2x)/(1-2sinxcosx)
=cos(2x)/[1-sin(2x)]
怎么证明cos2x/1-sin2x = cosx+sinx/cosx-sinx
Cosx+sinx分之1+sin2x+cos2x=2cos 证明
sinx×cos2x-sin2x×cosx
已知sinx=-1/2cosx,cosx-sinx/cosx+sinx+sin2x+cos2x的值
求证:(1-2sinx×cosx)/cos2x-sin2x=(cos2x-sin2x)/(1+2sinx×cosx)
证明1+sin2x/sinx+cosx=sinx+cosx(详细答案)
sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)
证明(cosx+sinX)^2=1+sin2X
解方程 sin2x/sinx=cos2x/cosx
cos2x+cosx=0,则sin2x+sinx
化简:(sinx+sin2x)/(1+cosx+cos2x)
已知1+sinx+cosx+sin2x+cos2x=0,求tanx