If $$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}},$$ then secθ is equal to:
A. $$\frac{{2{p^2}{q^2}}}{{{p^2} + {q^2}}}$$
B. $$\frac{1}{2}\left( {\frac{q}{p} + \frac{p}{q}} \right)$$
C. $$\frac{1}{{{p^2}}} + \frac{1}{{{q^2}}}$$
D. $$\frac{{{p^2}{q^2}}}{{{p^2} + {q^2}}}$$
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{{p^2}}}{{{q^2}}} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = \frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}} \cr & \sin \theta = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}} \cr} $$
$$\eqalign{ & AB = \sqrt {{{\left( {{p^2} + {q^2}} \right)}^2} - {{\left( {{p^2} - {q^2}} \right)}^2}} \cr & AB = \sqrt {4{p^2}{q^2}} \cr & AB = 2pq \cr & \sec \theta = \frac{{{p^2} + {q^2}}}{{2pq}} \cr & \sec \theta = \frac{1}{2}\left[ {\frac{p}{q} + \frac{q}{p}} \right] \cr} $$
This Question Belongs to Arithmetic Ability >> Trigonometry
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