Question

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} $

Answer 1

Answer: (B)

We have,
$\sin \theta=\sqrt{3} \cos \theta,-\pi<\theta<0$
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \theta=n \pi+\frac{\pi}{3}$ Put $n=-1$
$\therefore \theta=-\pi+\frac{\pi}{3}=\frac{-2 \pi}{3}$
$=-\frac{4 \pi}{6}$