Question

If $\sin \theta=\sqrt{3} \cos \theta,-\pi<\theta<0$, then $\theta$ is equal to

(A) $-\frac{5 x}{6}$

(B) $-\frac{4 \pi}{6}$

(C) $\frac{4 \pi}{6}$

(D) $\frac{5 \pi}{6}$

Answer 1

$\tan \theta=\sqrt{3}=\tan \frac{\pi}{3}$

$\Rightarrow \quad \theta=n \pi+\frac{\pi}{4}$

For $-\pi<\theta<0$ put $n=-1$, we get

$\theta=-\pi+\frac{\pi}{3}=\frac{-2 \pi}{3} \text { or } \frac{-4 \pi}{6}$

So, The correct option will be (B).