sinx/(1 - 1/cosx) - sinx/(1 + 1/cosx)sinx1−1cosx−sinx1+1cosx
sinx/((cosx - 1)/cosx) - sinx/((cosx + 1)/cosx)sinxcosx−1cosx−sinxcosx+1cosx
sinx xx cosx/(cosx - 1) - sinx xx cosx/(cosx + 1)sinx×cosxcosx−1−sinx×cosxcosx+1
(sinxcosx)/(cosx - 1) - (sinxcosx)/(cosx + 1)sinxcosxcosx−1−sinxcosxcosx+1
((sinxcosx)(cosx + 1))/((cosx - 1)(cosx + 1)) - ((sinxcosx)(cosx- 1))/((cosx + 1)(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)sinxcos2x+sinxcosxcos2x−1−sinxcos2x−sinxcosxcos2x−1
(sinxcos^2x + sinxcosx - sinxcos^2x + sinxcosx)/(cos^2x - 1)sinxcos2x+sinxcosx−sinxcos2x+sinxcosxcos2x−1
(2sinxcosx)/(cos^2x - 1)2sinxcosxcos2x−1
Applying the double angle identity 2sinxcosx = sin2x2sinxcosx=sin2x and the pythagorean identity cos^2x - 1 = -sin^2xcos2x−1=−sin2x we get the following:
(sin2x)/(-sin^2x) ->sin2x−sin2x→ 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!
