使【根号(1+sinθ/2)】/(1-sinθ/2)=secθ/2+tanθ/2成立的θ的取值范围
1个回答
cita=2k pai k=0,1,2-----n
如果对,
相关问题
secθ/√(1-sin2θ)+tanθ/√(csc2θ-1)= - 1 判断角θ所在的象限
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(
求证:(1-tanθ)/(1+tanθ)=(1-2sinθcosθ)/(cos2θ-sin2θ)
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
(tanθ-1)/(sinθ-cosθ)=secθ
求证:2(cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2)
已知1+sinθ根号下(1-cos^2θ)+cosθ根号下(1-sin^2θ)=0,则θ的取值范围是?
证明tan^θ-sin^2θ=tan^2θsin^2θ
试证:tanθ(1+sinθ)+sinθtanθ(1+sinθ)−sinθ=[tanθ+sinθ/tanθsinθ].
求证tan2θ-sin2θ=tan2θsin2θ