$$\left( {\frac{{{{\tan }^3}\theta }}{{{{\sec }^2}\theta }} + \frac{{{{\cot }^3}\theta }}{{{\text{cose}}{{\text{c}}^2}\theta }} + 2\sin \theta \cos \theta } \right)$$      ÷ (1 + cosec²θ + tan²θ), 0° < θ < 90°, is equal to:

Solution (By Examveda Team)

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