Q. If $\sin{\theta = \sqrt{3}\cos{\theta,\ \pi < \theta < 0}}$, then $\theta$ is equal to
A) $- \frac{5\pi}{6}$
B) $- \frac{4\pi}{6}$
C) $\frac{4\pi}{6}$
D) $\frac{5\pi}{6}$
Explanation
(b)
Given that, $\tan{\theta = \sqrt{3} = \tan{\frac{\pi}{3}\ \Rightarrow \ \theta = n\pi + \frac{\pi}{3}}}$
For $–\pi < \theta < 0\ put\ n = - 1,$ we get
$\theta = - \pi + \frac{\pi}{3} = \frac{- 2\pi}{3}\ or\ \frac{- 4\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.