[ 1/(cos^2x 乘以 √1+tan^2x) ] - √(1+sinx)/(1-sinx 其中 π/2 < x < π

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π/2 < x < π,所以 cosx<0;secx=1/袜稿稿cosx<0;sinx>告孝0

cos^2x * (√1+tan^2x) = cos^x*(-secx = -cosx
√敬雹[(1+sinx)/(1-sinx)] = √[(1+sinx)^2/cos^2x] = -(1+sinx)/cosx

所以[ 1/(cos^2x * √1+tan^2x) ] - √(1+sinx)/(1-sinx )
= -1/cosx+(1+sinx)/cosx
=tanx

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