$$\frac{{{{\left( {1 + \sec \theta \,{\text{cosec}}\,\theta } \right)}^2}{{\left( {\sec \theta - \tan \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \sec \theta } \right)}^2} + {{\left( {\cos \theta + {\text{cosec}}\,\theta } \right)}^2}}},$$        0° < θ < 90°, is equal to:

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{{\left( {1 + \sec \theta \,{\text{cosec}}\,\theta } \right)}^2}{{\left( {\sec \theta - \tan \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \sec \theta } \right)}^2} + {{\left( {\cos \theta + {\text{cosec}}\,\theta } \right)}^2}}} \cr & = \frac{{{{\left( {1 + \frac{1}{{\cos \theta \sin \theta }}} \right)}^2}{{\left( {\frac{1}{{\cos \theta }} - \frac{{\sin \theta }}{{\cos \theta }}} \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \frac{1}{{\cos \theta }}} \right)}^2} + {{\left( {\cos \theta + \frac{1}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{{{\left( {\frac{{\cos \theta \sin \theta + 1}}{{\cos \theta \sin \theta }}} \right)}^2}{{\left( {\frac{{1 - \sin \theta }}{{\cos \theta }}} \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\frac{{\cos \theta \sin \theta + 1}}{{\cos \theta }}} \right)}^2} + {{\left( {\frac{{\cos \theta \sin \theta + 1}}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{{{\left( {\frac{{\cos \theta \sin \theta + 1}}{{\cos \theta \sin \theta }}} \right)}^2}{{\left( {\frac{{1 - \sin \theta }}{{\cos \theta }}} \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\cos \theta \sin \theta + 1} \right)}^2}\left( {\frac{1}{{{{\cos }^2}\theta }} + \frac{1}{{{{\sin }^2}\theta }}} \right)}} \cr & = \frac{{{{\left( {\frac{{\cos \theta \sin \theta + 1}}{{\cos \theta \sin \theta }}} \right)}^2}{{\left( {\frac{{1 - \sin \theta }}{{\cos \theta }}} \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\cos \theta \sin \theta + 1} \right)}^2}\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right)}} \cr & = {\left( {\frac{{1 - \sin \theta }}{{\cos \theta }}} \right)^2}\left( {1 + \sin \theta } \right) \cr & = \frac{{{{\left( {1 - \sin \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\cos }^2}\theta }} \cr & = \frac{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)\left( {1 + \sin \theta } \right)}}{{\left( {1 - {{\sin }^2}\theta } \right)}} \cr & = \left( {1 + \sin \theta } \right) \cr} $$