1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程

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1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程

1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程
1-sin^2θ-cos^2θ/1-sinθ-cosθ
=1-1/1-sinθ-cosθ
=0
1-sin^6θ-cos^6θ/1-sin^4θ-cos^4θ
=1-(sin^6θ+cos^6θ)/1-sin^4θ-cos^4θ
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-1×(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-sin^4θ+sin^2θcos^2θ-cos^4θ]/1-sin^4θ-cos^4θ
=1 - (sin^2θcos^2θ/1-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ+cos^2θ-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ(1-sin^2θ)+cos^2θ(1-cos^2θ))
=1 - (sin^2θcos^2θ/sin^2θcos^2θ+cos^2θsin^2θ)
=1 - (sin^2θcos^2θ/2sin^2θcos^2θ)
=1-1/2
=1/2

sin^2θ+cos^2θ=1
1-sin^2θ-cos^2θ=1-（sin^2θ+cos^2θ）=0
所以1-sin^2θ-cos^2θ/1-sinθ-cosθ =0

0
(sinθ)^2+(cosθ)^2=1

化简：1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ 麻烦下大家 sinθ+cosθ=7/5,且tanθ>1, 则cos θ= 由于基础不好请过程详细一点!3qsinθ+cosθ=7/5 （sinθ+cosθ)^2 = 49/25 1+ 2sinθcosθ = 49/25 2sinθcosθ = 24/25 1-2sinθcosθ=1/25 （sinθ-cosθ)^2 = 1/25 sinθ-cosθ = 1/5 或 si 求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ 2sinθ-cosθ=1 (sinθ+cosθ+1)/(sinθ-cosθ+1)如题已知2sinθ-cosθ=1 求(sinθ+cosθ+1)/(sinθ-cosθ+1) 求证:（1+cosθ+cosθ/2） /（sinθ+sinθ/2）=sinθ/1-cosθ 1-cosθ/(2sinθ/2) 证明(1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)/(1-2sinθcosθ) 求证 (1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^θ-sin^2θ)/(1+2sinθcosθ) 求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以（1+2sinθcosθ） 2sinθ = cosθ + 1,怎么求sin 为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1) ((1+sinθ-cosθ)/(1+sinθ-cosθ))+ cot(θ/2) 化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot（θ/2） 化简2cosθ/√1-sin^2θ+√1-cos^2θ/sinθ （1+sinθ+cosθ）/（sin θ/2+cosθ/2） 2sinθcosθ+2sinθ-cosθ-1 = 0 0 sinθ-cosθ=1/2,则sin^3θ-cos^3θ=?. 已知sinθ-cosθ=1/2,求sin^3θ-cos^3θ