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If $$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}},$$ then secθ is equal to:
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}} \cr
& {\text{Apply componendo and dividendo}} \cr
& \frac{1}{{\sin \theta }} = \frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}} \cr
& \sin \theta = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}} \cr} $$

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

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