#sinx/(1 - 1/cosx) - sinx/(1 + 1/cosx)#
#sinx/((cosx - 1)/cosx) - sinx/((cosx + 1)/cosx)#
#sinx xx cosx/(cosx - 1) - sinx xx cosx/(cosx + 1)#
#(sinxcosx)/(cosx - 1) - (sinxcosx)/(cosx + 1)#
#((sinxcosx)(cosx + 1))/((cosx - 1)(cosx + 1)) - ((sinxcosx)(cosx- 1))/((cosx + 1)(cosx - 1))#
#(sinxcos^2x + sinxcosx)/(cos^2x - 1) - (sinxcos^2x - sinxcosx)/(cos^2x - 1)#
#(sinxcos^2x + sinxcosx - sinxcos^2x + sinxcosx)/(cos^2x - 1)#
#(2sinxcosx)/(cos^2x - 1)#
Applying the double angle identity #2sinxcosx = sin2x# and the pythagorean identity #cos^2x - 1 = -sin^2x# we get the following:
#(sin2x)/(-sin^2x) -># this is your answer.
*Beware: This is not the only possible answer; there are many ways of attacking this problem, with different answers completely within the realm of possibility. *
Hopefully this helps!
