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