The value of $$1 + \sqrt {\frac{{\cot \theta + \cos \theta }}{{\cot \theta - \cos \theta }}} ,$$    if 0° < θ < 90°, is equal to:

Solution (By Examveda Team)

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