A) \[-\frac{5\pi }{6}\]
B) \[-\frac{4\pi }{6}\]
C) \[\frac{4\pi }{6}\]
D) \[\frac{5\pi }{6}\]
Correct Answer: B
Solution :
We have, \[\sin \theta =\sqrt{3}\cos \theta ,\] \[-\pi <\theta <0\] \[\Rightarrow \] \[\tan \theta =\sqrt{3}\] \[\Rightarrow \] \[\theta =n\pi +\frac{\pi }{3}\] Put \[n=-1\] \[\therefore \] \[\theta =-\pi +\frac{\pi }{3}=\frac{-2\pi }{3}\] \[=-\frac{4\pi }{6}\]
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