若tanθ=2.则(sin2θ-cos2θ)/1+cos^2θ等于

若tanθ=2.则(sin2θ-cos2θ)/1+cos^2θ等于 若tan θ=3,求sin2θ -cos2θ 若tanθ=3,则sin2θ-cos2θ的值 若tan(π/4-θ)=3,则(cos2θ)/(1+sin2θ)等于? 若(1-tanθ)/(2+tanθ)=1 .则(cos2θ)/(1+sin2θ)的值为 (tanθ-1)/(2tanθ+1)=1,则cos2θ/(1+sin2θ)=? tanθ=2 sin2θ+2cos2θ 若tanθ=3,求sin2θ-cos2θ的值 证明:1+sin2θ+cos2θ/1+ sinθ-cos2θ=tanθ sin2θ=cos2θ+1,则cos2θ=? 若sin2θ=cos2θ+1,则cos2θ= 若tanβ=3,sin2β-cos2β= 已知1+sin2θ=-3cos2θ,且θ∈（0,∏/2）,则tanθ=( ) 求证：(3sin2θ-4cos2θ)/(2tanθ-1)-sin2θ=4cosθ^2 必修四数学 求证(sin2θ+1)/(sin2θ+cos2θ+1)=1/2(tanθ+1) sin2θ=tanθ-cos2θ,求sinθ- cosθ 求详解 tan(π/4+θ)=3,sin2θ-2cos2θ=? 已知tanθ=1/2,求sin2θ+cos2θ的值