Answer
$$\frac{{\sin \theta }}{{1 - \cos \theta }} - \frac{{\sin \theta \cos \theta }}{{1 + \cos \theta }} = \csc \theta \left( {1 + {{\cos }^2}\theta } \right)$$
Work Step by Step
$$\eqalign{
& \frac{{\sin \theta }}{{1 - \cos \theta }} - \frac{{\sin \theta \cos \theta }}{{1 + \cos \theta }} = \csc \theta \left( {1 + {{\cos }^2}\theta } \right) \cr
& {\text{We transform the more complicated left side to match the right side}}. \cr
& \frac{{\sin \theta }}{{1 - \cos \theta }} - \frac{{\sin \theta \cos \theta }}{{1 + \cos \theta }} = \frac{{\sin \theta \left( {1 + \cos \theta } \right) - \sin \theta \cos \theta \left( {1 - \cos \theta } \right)}}{{1 - {{\cos }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + \sin \theta {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \theta + \sin \theta {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sin \theta \left( {1 + {{\cos }^2}\theta } \right)}}{{{{\sin }^2}\theta }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sin \theta }}\left( {1 + {{\cos }^2}\theta } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \csc \theta \left( {1 + {{\cos }^2}\theta } \right) \cr
& {\text{Thus have verified that the given equation is an identity}} \cr} $$
