Answer
$$\left( {1 - {{\cos }^2}\alpha } \right)\left( {1 + {{\cos }^2}\alpha } \right) = 2{\sin ^2}\alpha - {\sin ^4}\alpha $$
Work Step by Step
$$\eqalign{
& \left( {1 - {{\cos }^2}\alpha } \right)\left( {1 + {{\cos }^2}\alpha } \right) = 2{\sin ^2}\alpha - {\sin ^4}\alpha \cr
& {\text{We transform the more complicated left side to match the right side}}. \cr
& \left( {1 - {{\cos }^2}\alpha } \right)\left( {1 + {{\cos }^2}\alpha } \right) = 1 - {\left( {{{\cos }^2}\alpha } \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\alpha } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {\left( {1 - {{\sin }^2}\alpha } \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \left( {1 - 2{{\sin }^2}\alpha + {{\sin }^4}\alpha } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 1 + 2{\sin ^2}\alpha - {\sin ^4}\alpha \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\sin ^2}\alpha - {\sin ^4}\alpha \cr
& {\text{Thus have verified that the given equation is an identity}} \cr} $$
