A) $ \frac{p^{2}+1}{p^{2}-1} $
B) $ \frac{p^{2}-1}{{{(p^{2}+1)}^{2}}} $
C) $ \frac{2p}{p^{2}+1} $
D) $ \frac{4p^{2}}{{{(p^{2}+1)}^{2}}} $
E) None of these
Correct Answer: C
$ \Rightarrow $ $ 1+\sin \theta =p\cos \theta $ (i) On squaring both sides, we get $ 1+{{\sin }^{2}}\theta +2\sin \theta =p^{2}{{\cos }^{2}}\theta $
$ \Rightarrow $ $ 1+1-{{\cos }^{2}}\theta +2\sin \theta =p^{2}{{\cos }^{2}}\theta $
$ \Rightarrow $ $ 2+2\sin \theta -{{\cos }^{2}}\theta =p^{2}{{\cos }^{2}}\theta $
$ \Rightarrow $ $ 2(1+\sin \theta )-{{\cos }^{2}}\theta =p^{2}{{\cos }^{2}}\theta $
$ \Rightarrow $ $ 2\times p\cos \theta -{{\cos }^{2}}\theta =p^{2}{{\cos }^{2}}\theta $ [from Eq.(i)]
$ \Rightarrow $ $ 2p\cos \theta =p^{2}{{\cos }^{2}}\theta +{{\cos }^{2}}\theta $
$ \Rightarrow $ $ 2p\cos \theta ={{\cos }^{2}}\theta (1+p^{2}) $
$ \therefore $ $ {{\cos }^{2}}\theta =\frac{2p\cos \theta }{(1+p^{2})} $
$ \Rightarrow $ $ \cos \theta =\frac{2p}{1+p^{2}} $
BETA
