tan（π/4＋θ）-tan（π/4+θ）=4，且-π＜θ＜-π/2，则sinθ=

tan（π/4＋θ）-tan（π/4+θ）=4，且-π＜θ＜-π/2，则sinθ=

写出过程

∵tan(π/4-θ)-tan(π/4+θ)=(tanπ/4 -tanθ)/(1+tanπ/4×tanθ)-(tanπ/4 +tanθ)/(1-tanπ/4×tanθ)

=(1-tanθ)/(1+tanθ)-(1+tanθ)/(1-tanθ)=-4tanθ/(1-tan²θ)=4

∴tan²θ-tanθ-1=0， ①

又∵-π＜θ＜-π/2，即tanθ＞0，sinθ＜0

故由①可知tanθ=(1+√5)/2

∴tan²θ=sin²θ/cos²θ=sin²θ/(1-sin²θ)=(3+√5)/2

即sin²θ=(3+√5)/(5+√5)，

所以sinθ=-√[(3+√5)/(5+√5)]=-√(50+10√5)/10

tan（π/4-θ)－tan（π/4+θ)＝(1-tanθ)/(1+tanθ)－(1+tanθ)/(1-tanθ)＝－4tanθ/[1-(tanθ)^2]＝4，所以(tanθ)^2－tanθ－1＝0，tanθ＝(1+√5)/2

-π＜θ＜-π/2，则sinθ＜0

sinθ＝－tanθ/√[1+(tanθ)^2]＝－√[(5+√5)/10]