化简(1-sin^2x)tanx
提问数学难题求证:sin^2x*tanx+cos^2x/tanx+2sinx*cosx=tanx+1/tanx
化简{sin(π+x)}/1+cos(3π-x)根号下(tanx-sinx)/(tanx+sinx)
求证(1-2sinxcosx)/(cos^2x-sin^2x)=(1-tanx)/(1+tanx)
求证1+2sinxcosx/cos^2x-sin^2x=1+tanx/1-tanx
证明:(1+2sinXcosX)/(sin^2X-cos^2X)=(tanX+1)/(tanX-1)
(cos^2x-sin^2x)/(1-2sinxcosx)=(1+tanx)/(1-tanx)
1-2sinx cosx /COS^2X-SIN^2X =1-tanx/1+tanx 求证
1).证明1-2sinxcosx/cos²x-sin²x=1-tanx/1+tanx
求证(1-2sinXcosX)/(cosX^2-sin^2X)=(1-tanX)/(1+tanX)
怎样把sinxcosx/sin^2xcos^2x化简为tanx/1+tanx?
已知sin(x-45°)=四分之根号2,求sinxcosx,tanx+1/tanx
sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)化简