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If $$\cos \theta = \frac{{2p}}{{1 + {p^2}}},$$ then tanθ is equal to:
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{ & \cos \theta = \frac{{2p}}{{1 + {p^2}}} \cr & A{B^2} = {\left( {1 + {p^2}} \right)^2} - {\left( {2p} \right)^2} \cr & AB = 1 - {p^2} \cr & \tan \theta = \frac{{1 - {p^2}}}{{2p}} \cr} $$
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