Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1\\
\dfrac{{1 - 2{{\cos }^2}\alpha }}{{1 + 2\sin \alpha .\cos \alpha }}.\dfrac{{\sin \alpha - cos\alpha }}{{\sin \alpha + \cos \alpha }}\\
= \dfrac{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) - 2{{\cos }^2}\alpha }}{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 2\sin \alpha .\cos \alpha }}.\dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}\\
= \dfrac{{{{\sin }^2}\alpha - {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha + 2.\sin \alpha .\cos \alpha + {{\cos }^2}\alpha }}.\dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}\\
= \dfrac{{\left( {\sin \alpha - \cos \alpha } \right).\left( {\sin \alpha + \cos \alpha } \right)}}{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2}}}.\dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}\\
= \dfrac{{{{\left( {\sin \alpha - \cos \alpha } \right)}^2}.\left( {\sin \alpha + \cos \alpha } \right)}}{{{{\left( {\sin \alpha + \cos \alpha } \right)}^3}}}\\
= {\left( {\dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}} \right)^2}
\end{array}\)
