已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急

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已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急已知

已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急
已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值
急

已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值急
由已知可得 tanα=tan[(α+π/4)-π/4]=[tan(α+π/4)-tan(π/4)] / [1+tan(α+π/4)tan(π/4)]
=(1/5-1)/(1+1/5)= -2/3 ,
所以 [sin(2α)-(sinα)^2] / [1-cos(2α)]
=sinα(2cosα-sinα) / [2(sinα)^2]
=(2cosα-sinα) / (2sinα)
=1/tanα-1/2
= -3/2-1/2
= -2 .

tan(π/4+α)=1/5
(1+tanα)/(1-tanα)=1/5
tanα=-2/3
(sin2α-sin²α)/(1-cos2α)
=(2sinαcosα-sin²α)/2sin²α
=(2tanα-tan²α)/2tan²α
=

tan(a+π/4)=[tana+tan(π/4)]/[1-tanatan(π/4)]=(tana+1)/(1-tana)=1/5
1-tana=5tana+5
6tana=-4
tana=-2/3
[sin(2a)-sin²a]/[1-cos(2a)]
=(2sinacosa-sin²a)/(2sin²a)
=(2cosa-sina)/(2sina)
=(2-tana)/(2tana)
=[2-(-2/3)]/[2(-2/3)]
=(8/3)/(-4/3)
=-2

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