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 $n=-1$
$\therefore \theta =-\pi +\frac{\pi }{3}=\frac{-2\pi }{3}$
$=-\frac{4\pi }{6}$