Answer:
Option A,C,D
Explanation:
Concept lnvolved. It is based on range of sin x i.e., [-1,1] and the internal for a< x< b.
Description of situation As $\theta, \phi \epsilon [0,2\pi]$ and
$\tan (2 \pi-\theta) 0,-1<\sin \theta <-\frac{\sqrt{3}}{2}$
$\tan (2 \pi-\theta) >0 $ $ \Rightarrow$ $-\tan \theta >0$
$\therefore$ $\theta \epsilon $ II or IV quadrant
also, $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$
$\Rightarrow$ $\frac{4 \pi}{3} < \theta < \frac{5 \pi}{3}$ but $\theta \epsilon$ II or IV quadrant
$\Rightarrow$ $\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$ .....(i)
Sol. here
$2 \cos \theta (1-\sin \phi )$
=$ \sin^{2} \theta (\tan \frac{ \theta}{2}+\cot \frac{\theta}{2}) \cos \phi-1$
$\Rightarrow$ $2 \cos \theta-2 \cos \theta \sin \phi$
= $\sin^{2} \theta\left(\frac{\sin^{2}\frac{\theta}{2}+\cos^{2} \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1$
=$2 \cos \theta-2 cos \theta \sin \phi $
=$2\sin^{2} \theta\left(\frac{1}{\sin\theta}\right) \cos \phi-1$
=$2 \cos \theta +1=2 \sin \phi \cos \theta+2 \sin \theta \cos \phi$
$\Rightarrow$ $2 \cos \theta+1=2 \sin ( \theta+\phi)$.....................(ii)
From Eq.(i)
$\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$ $\Rightarrow 2 \cos \theta+1 \epsilon(1,2)$
$\therefore$ $1 < \sin (\theta+\phi)<2$
$\rightarrow$ $\frac{1}{2} < \sin (\theta +\phi) <1$......(iii)
$\Rightarrow$ $\frac{\pi}{6} < \theta +\phi < \frac{5 \pi}{6}$ or $\frac{13 \pi}{6} < \theta +\phi < \frac{17 \pi}{6}$
$\therefore$ $\frac{\pi}{6} -\theta < \phi < \frac{5 \pi}{6}- \theta$
or $ \frac{13 \pi}{6}-\theta < \phi < \left(\frac{17 \pi}{6}\right)-\theta$
$\Rightarrow$ $\phi \epsilon\left(-\frac{3 \pi}{2},-\frac{2\pi}{3}\right)or \left(\frac{2\pi}{3},\frac{7 \pi}{6}\right),$
as $\theta \epsilon\left(\frac{3 \pi}{2},\frac{5 \pi}{3}\right)$
