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