sinx/(1-cosx)*根号[（tanx-sinx)/(tanx+sinx)]

√[（tanx-sinx)/(tanx+sinx)]
=√1-cosx/1+cosx
=√((1-cosx)^2/sin^2
=(1-cosx)/sinx
所以sinx/(1-cosx)*根号[（tanx-sinx)/(tanx+sinx)]=sinx/(1-cosx)*(1-cos)/sinx=1