Question

Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0$ and $-1<\sin \theta<-\frac{\sqrt{3}}{2} .$ Then $\varphi$ cannot satisfy

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

Answer 1

Answer: (A), (C), (D)

$2 \cos \theta(1-\sin \varphi)=\frac{2 \sin ^{2} \theta}{\sin \theta} \cos \varphi-1=2 \sin \theta \cos \varphi-1 $
$2 \cos \theta-2 \cos \theta \sin \varphi=2 \sin \theta \cos \varphi-1 $
$2 \cos \theta+1=2 \sin (\theta+\varphi) $
$\tan (2 \pi-\theta)>0 \Rightarrow \tan \theta<0 \text { and }-1<\sin \theta<-\frac{\sqrt{3}}{7} $
$\Rightarrow \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) $
$\frac{1}{2}<\sin (\theta+\varphi)<1 $
$\Rightarrow 2 \pi+\frac{\pi}{6}<\theta+\varphi<\frac{5 \pi}{6}+2 \pi $
$2 \pi+\frac{\pi}{6}-\theta_{\max }<\varphi<2 \pi+\frac{5 \pi}{6}-\theta_{\min }$
$\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$