<s>[INST] Finding the trigonometrical solution within the specified region Let $\theta ,\varphi  \in \left[ {0,2\pi } \right]$ be such that
$2\cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\tan \frac{\theta }{2} + \cot \frac{\theta }{2}} \right)\cos \varphi  - 1$
$ - 1 < \sin \theta  <  - \frac{{\sqrt 3 }}{2}$ & $\tan \left( {2\pi  - \theta } \right) > 0$, 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) $\frac{3\pi}{2}<\varphi<2\pi$
The official answer is ACD.
My approach is as follow
$ - 1 < \sin \theta  <  - \frac{{\sqrt 3 }}{2} \Rightarrow \theta  \in \left( {\frac{{4\pi }}{3},\frac{{3\pi }}{2}} \right) \cup \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right)$
$\tan \left( {2\pi  - \theta } \right) > 0 \Rightarrow \tan \theta  < 0 \Rightarrow \theta  \in \left( {\frac{\pi }{2},\pi } \right) \cup \left( {\frac{{3\pi }}{2},2\pi } \right)$
Hence $\theta  \in \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right)$
$2\cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\tan \frac{\theta }{2} + \cot \frac{\theta }{2}} \right)\cos \varphi  - 1$
$ \Rightarrow \cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\frac{{{{\sin }^2}\frac{\theta }{2} + {{\cos }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)\cos \varphi  - \frac{1}{2} \Rightarrow \cos \theta \left( {1 - \sin \varphi } \right) = \sin \theta \cos \varphi  - \frac{1}{2}$
$ \Rightarrow \frac{1}{2} + \cos \theta  - \sin \varphi \cos \theta  = \sin \theta \cos \varphi  \Rightarrow \frac{1}{2} + \cos \theta  = \sin \theta \cos \varphi  + \sin \varphi \cos \theta  \Rightarrow \frac{1}{2} + \cos \theta  = \sin \left( {\theta  + \varphi } \right) = T$
$\theta  \in \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right) \Rightarrow \frac{{dT}}{{d\theta }} =  - \sin \theta $
Therefore T is increasing
Upto to this step I am satisfied with my approach
But from here how do I match $\sin \theta \cos \varphi  + \sin \varphi \cos \theta  \Rightarrow \frac{1}{2} + \cos \theta  = \sin \left( {\theta  + \varphi } \right) = T$ with the given answer and arrive at the proper result
 [/INST] Since $\theta \in (3\pi/2, 5\pi/3)$ is in 4th quadrant. You must have $\frac 12+\cos \theta \gt 0\implies \sin (\theta +\phi)\gt 0\tag 1$ 
In case $(b),\theta +\phi \in (\pi/2+3\pi/2,4\pi/3+5\pi/3)=(2\pi, 3\pi)$ and that is correct as it satisfies $(1)$. 
In case $(a),\theta+\phi \in (3\pi/2, 5\pi/3+\pi/2)=(3\pi/2, 13\pi/6)$ is not correct as $\theta +\phi=2\pi$, for example, will create problem for $(1)$. 
Similarly rule out other possibilities.
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