设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(11t^2)设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2)

设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(11t^2)设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2) 设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2)sec没有学过 设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2) 求证tan^2θ-sin^2θ=tan^2θsin^2θ 设cosθ=t(t≥0),求sinθ和tanθ的值 求证 tanθ(1+sinθ )+sinθ /tanθ (1+sinθ )-sinθ =tanθ+sinθ/tanθsinθ 高中三角函数证明y=-t^2/2 + 1/2设t=tanθ及y=[(cos^2θ-sin^2θ)/2sinθcosθ ] * tanθ证明y=-t^2/2 + 1/2 求证2(cosθ -sinθ )/1+sinθ +cosθ =tan(π/4-θ /2)-tanθ /2 求证：2（cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2) 已知tanθ+sinθ=a,tanθ-sinθ=b,求证(a2-b2)2=16ab 求证：tan²θ-sin²θ=tan²θ·sin²θ 已知(sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=1,求证（tana)^2÷（tan)^2=（sinγ)^2设cosβ=t（t≥0），秋sinβ和tanβ 数学的求证题已知tanθ+sinθ=a,tanθ-sinθ=b求证：（a^2-b^2)^2=16ab 已知(sin^2α/sin^2β)+cos^2αcos^2θ=1,求证tan^2α=sin^2θtan^2β 已知tan^2θ=2tan^2a+1,求证:cos2θ+sin^2a=0 已知tan^2θ=2tan^2a+1,求证:cos2θ+sin^2a=0 tan²α=2tan²θ+1求证cos2α+sin²θ=0 cosθ=t,求sinθ,tanθ