$\begin{array}{l} \dfrac{2}{{\sin \alpha }} - \dfrac{{\sin \alpha }}{{1 + \cos \alpha }} = \dfrac{{2\left( {1 + \cos \alpha } \right) - {{\sin }^2}\alpha }}{{\left( {1 + \cos \alpha } \right).\sin \alpha }} = \dfrac{{2 + 2\cos \alpha - {{\sin }^2}\alpha }}{{\left( {1 + \cos \alpha } \right)\sin \alpha }}\\ = \dfrac{{2\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 2\cos \alpha - {{\sin }^2}\alpha }}{{\left( {1 + \cos \alpha } \right)\sin \alpha }}\\ = \dfrac{{2{{\cos }^2}\alpha + 2\cos \alpha + {{\sin }^2}\alpha }}{{\left( {1 + \cos \alpha } \right)\sin \alpha }} = \dfrac{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {{\cos }^2}\alpha + 2\cos \alpha }}{{\left( {1 + \cos \alpha } \right)\sin \alpha }}\\ = \dfrac{{{{\cos }^2}\alpha + 2\cos \alpha + 1}}{{\left( {1 + \cos \alpha } \right)\sin \alpha }} = \dfrac{{{{\left( {\cos \alpha + 1} \right)}^2}}}{{\left( {1 + \cos \alpha } \right)\sin \alpha }} = \dfrac{{1 + \cos \alpha }}{{\sin \alpha }} \end{array}$
