tanx=6,1/2sin^2x+1/3cos^2x=?
提问数学难题求证:sin^2x*tanx+cos^2x/tanx+2sinx*cosx=tanx+1/tanx
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
(1+tanx)/(1-tanx)=3+2根号2,求(sin x+cosx)^2-(cos^3x)/sinx
已知(1+tanx)/(1-tanx)=3+根号二,求cos²x+sinxcosx+2sin²x的值
若tanx/tanx-1=-1,求sin(π/2+x)cos(3π-x)的值
已知tanx=1/2,求sin x-cos x/2sinx+3cosx的值
证明1/sin^x+1/cos^2x-1/tanx^2x=2+tan^2