tan²α=2tan²θ+1求证cos2α+sin²θ=0
已知tan²β=2tan²α+1,求证cos2β+sin²α=0
tan^2θ =2tan^2Ф+1 求证 cos2θ +sin^2Ф=0
已知tan^2θ=2tan^2a+1,求证:cos2θ+sin^2a=0
已知tanα=-1/2,则(sinα-cosα)²/cos2α=?
求证:1+2sinαcosα/cos2α-sin2α=1+tanα/1-tanα
已知tan[α+π/4]=2,求cos2α+3sin²α的值
tan^2α=2tan^2 β+1则 cos2α+sin^2β等于?
tanα=根号2,则1-sin²α分之cos²α-cos2α的值
sin²2α+sin2α×cosα-cos2α=1,α∈(0,π/2),求sinα,tanα的值
已知sin²2α+sin2αcosα-cos2α=1α∈(0,π/2)求sinα,tanα的值
已知tan^2θ=2tan^2α+1,求cos2α-2cos2θ的值
求证:(1-tanθ)/(1+tanθ)=(1-2sinθcosθ)/(cos2θ-sin2θ)