$$\frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} = ?$$

$$\eqalign{ & \frac{{{{\left( {1 + \cos \theta } \right)}^2} + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{1 + {{\cos }^2}\theta + 2\cos \theta + {{\sin }^2}\theta }}{{\left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right){{\sin }^2}\theta }} \cr & = \frac{{2\left( {1 + \cos \theta } \right)}}{{\frac{{\left( {1 - {{\sin }^2}\theta } \right)}}{{{{\sin }^2}\theta }}.{{\sin }^2}\theta }} \cr & = \frac{{2\left( {\cos \theta + 1} \right)}}{{{{\cos }^2}\theta }} \cr & = 2\sec \theta \left( {\frac{{\cos \theta }}{{\cos \theta }} + \frac{1}{{\cos \theta }}} \right) \cr & = 2\sec \theta \left( {1 + \sec \theta } \right) \cr} $$