If 1 + 1 - = p^2q^2, then secθ is equal to:

Answer: Option B

$$\begin{array}{rl}& \frac{1+\sin \theta }{1-\sin \theta }=\frac{p^2}{q^2}\\ & \text{Apply componendo and dividendo}\\ & \frac{1}{\sin \theta }=\frac{p^2+q^2}{p^2-q^2}\\ & \sin \theta =\frac{p^2-q^2}{p^2+q^2}\end{array}$$

$$\begin{array}{rl}& AB=\sqrt{{\left(p^2+q^2\right)}^2-{\left(p^2-q^2\right)}^2}\\ & AB=\sqrt{4p^2{q}^2}\\ & AB=2pq\\ & \sec \theta =\frac{p^2+q^2}{2pq}\\ & \sec \theta =\frac{1}{2}\left[\frac{p}{q}+\frac{q}{p}\right]\end{array}$$