A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
\[\sec \theta +\tan \theta =\sqrt{3}\] ?.(i) Also we have \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] ?..(ii) \[\Rightarrow \] \[\sec \theta -\tan \theta =\frac{1}{\sqrt{3}}\] ?..(iii) Now (i) and (iii) gives, \[\tan \theta =\frac{1}{2}\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)=\frac{1}{\sqrt{3}}=\tan \left( \frac{\pi }{6} \right)\] \[\Rightarrow \] \[\theta =n\pi +\frac{\pi }{6}\]. \[\therefore \] Solutions for \[0\le \theta \le 2\pi \] are \[\frac{\pi }{6}\] and \[\frac{7\pi }{6}\]. Hence there are two solutions.
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