f(x)=[1+sin²x+sin(π-x)-cos²(π-x)]/[2sin(π+x)cos(π-x)-cos(π+x)]
=(1+sin²x+sinx-cos²x)/[2(-sinx)(-cosx)-(-cosx)]
=(sin²x+cos²x+sin²x+sinx-cos²x)/(2sinxcosx+cosx)
=[sinx(2sinx+1)/cosx(2sinx+1)]
=tanx
故
f[-(23/6)π]
=f(-4π+π/6)
=tan(-4π+π/6)
=tan(π/6)
=√3/3