sin2x+2sin²x/1-tanx
=sin2x+2sin²x-1+1/1-tanx
=(sin2x-cos2x+1)/(1-tanx)
=(2sinx*cosx-cosx^2+sinx^2+1)/(1-tanx)
=[(sinx+cosx)^2-(cosx+sinx)(cosx-sinx)]cosx/(cosx-sinx)
=2sinx*cosx(sinx+cosx)/(cosx-sinx)
又∵17π/12＜x＜7π/4,∴5π/3＜x+π/4＜2π
∴sinx＜0,sin（x+π/4)＜0
又∵cosx=cos[(π/4+x)-π/4]
=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4
=3/5*√2/2+(-4/5)*√2/2
=-√2/10
∴sinx=-7√2/10
∴原式=2*(-√2/10)*(-7√2/10)(-8√2/10)/(6√2/10)
=-28/75