(sec∅ - tan∅)²(1 + sin∅)² ÷ cos²∅ = ?

Solution(By Examveda Team)

$$\eqalign{ & {\left( {\sec \phi - \tan \phi } \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{1}{{\cos \phi }} - \frac{{\sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = {\left( {\frac{{1 - \sin \phi }}{{\cos \phi }}} \right)^2}{\left( {1 + \sin \phi } \right)^2} \div {\cos ^2}\phi \cr & = \frac{{{{\left( {1 - \sin \phi } \right)}^2}{{\left( {1 + \sin \phi } \right)}^2}}}{{{{\cos }^2}\phi }} \times \frac{1}{{{{\cos }^2}\phi }} \cr & = \frac{{{{\left( {1 - {{\sin }^2}\phi } \right)}^2}}}{{{{\cos }^4}\phi }} \cr & = \frac{{{{\cos }^4}\phi }}{{{{\cos }^4}\phi }} \cr & = 1 \cr} $$