A) \[\frac{\pi }{6}\]
B) \[\frac{5\pi }{6}\]
C) \[\frac{7\pi }{6}\]
D) \[-\frac{\pi }{6}\]
Correct Answer: B
Solution :
\[\tan \theta =\frac{-1}{\sqrt{3}}=\tan \left( \pi -\frac{\pi }{6} \right)\], \[\sin \theta =\frac{1}{2}=\sin \left( \pi -\frac{\pi }{6} \right)\] and \[\cos \theta =\frac{-\sqrt{3}}{2}\]= \[\cos \left( \pi -\frac{\pi }{6} \right)\] Hence principal value is \[\theta =\frac{5\pi }{6}\].
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