A) 5/4
B) 4/6
C) $ \sqrt{3}/2 $
D) ½
Correct Answer: B
$ \Rightarrow $ $ (r\cos \theta -\sqrt{3})=0 $
$ \Rightarrow $ $ r\cos \theta =\sqrt{3} $ (i)
$ \Rightarrow $ $ {{(r\sin \theta -1)}^{2}}=0 $
$ \Rightarrow $ $ (r\sin \theta -1)=0 $
$ \Rightarrow $ $ r\sin \theta =1 $
(ii)
On squaring Eqs. (i) and (ii) and adding,
$ r^{2}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=3+1=4 $
$ \Rightarrow $ $ r=2 $
$ \cos \theta =\frac{\sqrt{3}}{2} $
$ \Rightarrow $ $ \theta =\frac{\pi }{6} $
$ \Rightarrow $ $ \sin \theta =\frac{1}{2} $
$ \Rightarrow $ $ \theta =\frac{\pi }{6} $
$ \frac{r\tan \theta +\sec \theta }{r\sec \theta +\tan \theta }=\frac{2\tan \pi /6+\sec \pi /6}{2\sec \pi /6+\tan \pi /6} $
$ =,\frac{2\times \frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}}{2\times \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}}=\frac{\frac{4}{\sqrt{3}}}{\frac{5}{\sqrt{3}}}=\frac{4}{5} $
BETA
