Question

Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1<\sin \theta<-\frac{\sqrt{3}}{2}$. Then value of $\phi$ is

• A. $0<\phi<\frac{\pi}{2}$
• B. $\frac{\pi}{2}<\phi<\frac{4 \pi}{3}$
• C. $\frac{4 \pi}{3}<\phi<\frac{3 \pi}{2}$
• D. $\frac{3 \pi}{2}<\phi<2 \pi$

Answer 1

Answer: (B)

$\text { As } \tan (2 \pi-\theta)>0,-1<\sin \theta<-\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]$
$\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3}$
Now $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta(\tan \theta / 2+\cot \theta / 2) \cos \phi-1$
$\Rightarrow 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1$
$ \Rightarrow 2 \cos \theta+1=2 \sin (\theta+\phi)$
As $\theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) $
$\Rightarrow 2 \cos \theta+1 \in(1,2)$
$\Rightarrow 1<2 \sin (\theta+\phi)< 2$
$ \Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1$
As $\theta+\phi \in[0,4 \pi]$
$\Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) $
$\text { or } \theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) $
$\Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta $
$\text { or } \frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta$
$\Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) \left(\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right)$