高一数学题.化简[√(1+sinα)/(
化简[√(1+sinα)/(1-sinα)-√(1-sinα)/(1+sinα)]·[√（1+sinα)/(1-cosα)-√(1-cosα)/(1+cosα)]
根据对称性，倒数第二个分式的分子是否应该为1+cosα？
[√(1+sinα)/(1-sinα)-√(1-sinα)/(1+sinα)]*[√(1+cosα)/(1-cosα)-√(1-cosα)/(1+sinα)]
显然，sinα、cosα≠±1
对于上式中每一个根号有如下：
√(1+sinα)/(1-sinα)=√[(1+sinα)^2/(1-sinα)(1+sinα)^2]
=√[(1+sinα)^2/(1-sin^2 α...全部
化简[√(1+sinα)/(1-sinα)-√(1-sinα)/(1+sinα)]·[√（1+sinα)/(1-cosα)-√(1-cosα)/(1+cosα)]
根据对称性，倒数第二个分式的分子是否应该为1+cosα？
[√(1+sinα)/(1-sinα)-√(1-sinα)/(1+sinα)]*[√(1+cosα)/(1-cosα)-√(1-cosα)/(1+sinα)]
显然，sinα、cosα≠±1
对于上式中每一个根号有如下：
√(1+sinα)/(1-sinα)=√[(1+sinα)^2/(1-sinα)(1+sinα)^2]
=√[(1+sinα)^2/(1-sin^2 α)]
=√[(1+sinα)^2/cos^2 α]
=|1+sinα|/|cosα|
=(1+sinα)/|cosα|
√(1-sinα)/(1+sinα)=1/[√(1+sinα)/(1-sinα)]=|cosα|/(1+sinα)
√(1+cosα)/(1-cosα)=√[(1+cosα)^2/(1-cosα)(1+cosα)]
=√[(1+cosα)^2/(1-cos^2 α)]
=√[(1+cosα)^2/sin^2 α]
=|1+cosα|/|sinα|
=(1+cosα)/|sinα|
√(1-cosα)/(1+cosα)=1/[√(1+cosα)/(1-cosα)]=|sinα|/(1+cosα)
所以：
原式=(1+sinα)/|cosα|+|cosα|/(1+sinα)+(1+cosα)/|sinα|+|sinα|/(1+cosα)
=[(1+sinα)^2+|cosα|^2]/[(1+sinα)*|cosα|]+[(1+cosα)^2+|sinα|^2]/[(1+cosα)*|sinα|]
=[1+2sinα+(sin^2 α+cos^2 α)]/[(1+sinα)*|cosα|]+[1+2cosα+(cos^2 α+sin^2 α)]/[(1+cosα)*|sinα|]
=[2(1+sinα)]/[(1+sinα)*|cosα|]+[2(1+cosα)]/[(1+cosα)*|sinα|]
=2/|cosα|+2/|sinα|。
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