QJEEAdv12Maa550: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Maa550 Solution: Let be such that and then cannot satisfy

Given, \tan (2 \pi-\theta)>0

$\begin{aligned}& \Rightarrow 0<(2 \pi-\theta)<\frac{\pi}{2} \text { or } \pi<(2 \pi-\theta)<\frac{3 \pi}{2} \\& \Rightarrow \frac{3 \pi}{2}<\theta<2 \pi \text { or } \frac{\pi}{2}<\theta<\pi \quad \text{... (i)}\end{aligned}$

Also, -1<\sin \theta<\frac{-\sqrt{3}}{2}

\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3} \quad \text{... (ii)}

From (i) and (ii),

$\Rightarrow \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\quad \text{... (iii)}$

Now,

$\begin{aligned}& 2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1 \\& \Rightarrow 2 \cos \theta(1-\sin \phi) \\& =\sin ^2 \theta\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right) \cos \phi-1 \\& \Rightarrow 2 \cos \theta(1-\sin \phi) \\& =\sin ^2 \theta\left(\frac{\sin ^2 \frac{\theta}{2}+\cos ^2 \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1\end{aligned}$

$\begin{aligned}& \Rightarrow \quad 2 \cos \theta(1-\sin \phi) \\&=\sin \theta \cdot \sin \theta\left(\frac{1}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1\end{aligned}$

$\begin{aligned}& \Rightarrow \quad 2 \cos (1-\sin \phi)=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cdot \cos \phi \cdot \sin \theta}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}-1 \\& \Rightarrow \quad 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1 \\& \Rightarrow \quad 2 \cos \theta+1=2 \sin (\theta+\phi) \\& \text { As } \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) \quad \text{[from (iii)]}\\& \Rightarrow 2 \cos \theta+1 \in(1,2) \\& \Rightarrow 1<2 \sin (\theta+\phi)<2 \\& \Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1\end{aligned}$

$\begin{aligned}& \text { Now, } \theta, \phi \in[0,2 \pi] \\& \theta+\phi \in[0,4 \pi] \\& \Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) \text { or } \theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) \\& \Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta \text { or } \frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta \\& \Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) \\& {\left[\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right] }\end{aligned}$