求证sinx-cosx+1/sinx+cosx-1=1+sinx/cosx
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(sinx-cosx+1)/(sinx+cosx-1)
=[2sin(x/2)cos(x/2)+2(sin(x/2))^2] / [ 2sin(x/2)cos(x/2)-2(sin(x/2))^2
=[sin(x/2) +cos(x/2)] /[cos(x/2) -sin(x/2)]
(1+sinx)/cosx=(sin(x/2)+cos(x/2))^2/[cos(x/2))^2-sin(x/2)^2]
=[sin(x/2)+cos(x/2)]/[cos(x/2)-sin(x/2)]
左=右
=[2sin(x/2)cos(x/2)+2(sin(x/2))^2] / [ 2sin(x/2)cos(x/2)-2(sin(x/2))^2
=[sin(x/2) +cos(x/2)] /[cos(x/2) -sin(x/2)]
(1+sinx)/cosx=(sin(x/2)+cos(x/2))^2/[cos(x/2))^2-sin(x/2)^2]
=[sin(x/2)+cos(x/2)]/[cos(x/2)-sin(x/2)]
左=右
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