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If $$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}},$$    then secθ is equal to:

If $$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}},$$    then secθ is equal to: Correct Answer $$\frac{1}{2}\left( {\frac{q}{p} + \frac{p}{q}} \right)$$

$$\eqalign{ & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = \frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}} \cr & \sin \theta = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}} \cr} $$
Trigonometry mcq question image
$$\eqalign{ & AB = \sqrt {{{\left( {{p^2} + {q^2}} \right)}^2} - {{\left( {{p^2} - {q^2}} \right)}^2}} \cr & AB = \sqrt {4{p^2}{q^2}} \cr & AB = 2pq \cr & \sec \theta = \frac{{{p^2} + {q^2}}}{{2pq}} \cr & \sec \theta = \frac{1}{2}\left \cr} $$

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