The expression $$\frac{{{{\cos }^4}\theta - {{\sin }^4}\theta + 2{{\sin }^2}\theta + 3}}{{\left( {{\text{cosec}}\,\theta + \cot \theta + 1} \right)\left( {{\text{cosec}}\,\theta - \cot \theta + 1} \right) - 2}},$$         is equal to:

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{{\cos }^4}\theta - {{\sin }^4}\theta + 2{{\sin }^2}\theta + 3}}{{\left( {{\text{cosec}}\,\theta + \cot \theta + 1} \right)\left( {{\text{cosec}}\,\theta - \cot \theta + 1} \right) - 2}} \cr & = \frac{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2{{\sin }^2}\theta + 3}}{{\left( {{\text{cosec}}\,\theta + 1 + \cot \theta } \right)\left( {{\text{cosec}}\,\theta + 1 - \cot \theta } \right) - 2}} \cr & = \frac{{\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2{{\sin }^2}\theta + 3}}{{{{\left( {{\text{cosec}}\,\theta + 1} \right)}^2} - {{\cot }^2}\theta - 2}} \cr & = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta + 3}}{{{\text{cose}}{{\text{c}}^2}\theta + 1 + 2{\text{cosec}}\,\theta - {{\cot }^2}\theta - 2}} \cr & = \frac{4}{{2{\text{cosec}}\,\theta }} \cr & = 2\sin \theta \cr} $$