(sec∅ - tan∅)2(1 + sin∅)2 ÷ cos2∅ = ?

Answer: Option B

$$\begin{array}{rl}& {\left(\sec \varphi -\tan \varphi \right)}^2{\left(1+\sin \varphi \right)}^2÷{\cos }^2\varphi \\ & ={\left(\frac{1}{\cos \varphi }-\frac{\sin \varphi }{\cos \varphi }\right)}^2{\left(1+\sin \varphi \right)}^2÷{\cos }^2\varphi \\ & ={\left(\frac{1-\sin \varphi }{\cos \varphi }\right)}^2{\left(1+\sin \varphi \right)}^2÷{\cos }^2\varphi \\ & =\frac{{\left(1-\sin \varphi \right)}^2{\left(1+\sin \varphi \right)}^2}{{\cos }^2\varphi }\times \frac{1}{{\cos }^2\varphi }\\ & =\frac{{\left(1-{\sin }^2\varphi \right)}^2}{{\cos }^4\varphi }\\ & =\frac{{\cos }^4\varphi }{{\cos }^4\varphi }\\ & =1\end{array}$$