The expression (tanθ + cotθ)(secθ + tanθ)(1 - sinθ), 0° < θ < 90° is equal to:

Solution (By Examveda Team)

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