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