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