已知 -π/2<x<0,sinx+cosx=1/5.求(sin2x+2sin平方x)/(1-tanx)的值

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由(sinx)^2+(cosx)^2=1
sinx+cosx=1/5
且sinx<0
解得sinx=-3/5,cosx=4/5
tgx=sinx/cosx=-3/4,sin2x=2sinxcosx=-24/袜晌橘李25
(sin2x+2(sinx)^2)/(1-tgx)
=(-24/25+2*9/25)/(1+3/告伍锋4)
=-24/175

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