已知2sinθ-cosθ=1,则(sinθ+cosθ+1)/(sinθ-cosθ+1)
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
已知tanθ=根号2,求(1)(cosθ+sinθ)/(cosθ-sinθ);(2)sin²θ-sinθcos
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
已知θ为第三象限角,1-sinθcosθ-3cos^2=0则5sin^2θ+3sinθcosθ=?
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
sinθ-cosθ=1/2,则sin^3θ-cos^3θ=?.
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcos
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
已知sin^4θ+cos^4θ=1,求sinθ+cosθ的值
若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ