The expression $$\frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1,$$       0° < θ < 90°, equals:

Solution (By Examveda Team)

$$\eqalign{ & \frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1 \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \frac{{\left( {1 - 2 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right)\left( {1 + 1} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {1 - 2 \times \frac{1}{4}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{4} + \frac{1}{4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{2}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{2}} \right)\left( 2 \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 }} - 1 \cr & = \frac{1}{2} - 1 \cr & = - \frac{1}{2} \cr & {\text{Option C is answer}} \cr & \Rightarrow - {\sin ^2}\theta \cr & = - {\sin ^2}{45^ \circ } \cr & = - {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = - \frac{1}{2} \cr} $$