( 1 + )^2 + ^2 ( cose c^2 - 1 ) ^2 = ?

Answer: Option D

$$\begin{array}{rl}& \frac{{\left(1+\cos \theta \right)}^2+{\sin }^2\theta }{\left(\text{cose}{\text{c}}^2\theta -1\right){\sin }^2\theta }\\ & =\frac{1+{\cos }^2\theta +2\cos \theta +{\sin }^2\theta }{\left(\text{cose}{\text{c}}^2\theta -1\right){\sin }^2\theta }\\ & =\frac{2\left(1+\cos \theta \right)}{\frac{\left(1-{\sin }^2\theta \right)}{{\sin }^2\theta }.{\sin }^2\theta }\\ & =\frac{2\left(\cos \theta +1\right)}{{\cos }^2\theta }\\ & =2\sec \theta \left(\frac{\cos \theta }{\cos \theta }+\frac{1}{\cos \theta }\right)\\ & =2\sec \theta \left(1+\sec \theta \right)\end{array}$$