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