Question

Let $\theta ,\varphi \in [0,2\pi ]$ be such that $2\cos \theta (1-\sin \varphi )={\sin }^2\theta (\tan \frac{\theta }{2}+\cot \frac{\theta }{2})\cos \varphi -1,\tan (2\pi -\theta )>0$ 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)

$\frac{3\pi }{2}<\varphi <2\pi$


$\tan (2\pi -\theta )>0$
$0<2\pi -\theta <\frac{\pi }{2}$ or $\pi <2\pi -\theta <\frac{3\pi }{2}$
$\Rightarrow \frac{3\pi }{2}<\theta <2\pi$ or $\frac{\pi }{2}<\theta <\pi \dots \left(1\right)$

Also $-1<\sin \theta <-\frac{\sqrt{3}}{2}$
$\frac{3\pi }{2}<\theta <\frac{5\pi }{3}\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\theta \in (\frac{3\pi }{2},\frac{5\pi }{3})\dots \left(3\right)$

Now,
$2\cos \theta (1-\sin \varphi )={\sin }^2\theta (\tan \frac{\theta }{2}+\cot \frac{\theta }{2})\cos \varphi -1$
$\Rightarrow \cos \theta +\frac{1}{2}=\sin (\theta +\varphi )$
$\Rightarrow 2\cos \theta +1=2\sin (\theta +\varphi )\dots \left(4\right)$

As $\theta \in (\frac{3\pi }{2},\frac{5\pi }{3})\Rightarrow 2\cos \theta +1\in (1,2)$
$\cos \theta \in (0,\frac{1}{2})$
$\sin (\theta +\varphi )\in (\frac{1}{2},1)\cdots \left(5\right)$

As $\theta +\varphi \in [0,4\pi ]\cdots \left(6\right)\text{{Given}}$

From equation $\left(5\right)$ and $\left(6\right)$

$\theta +\varphi \in (\frac{\pi }{6},\frac{5\pi }{6})\text{or}\theta +\varphi \in (\frac{13\pi }{6},\frac{17\pi }{6})$

$\because \theta \in (\frac{3\pi }{2},\frac{5\pi }{3})$
$\Rightarrow \varphi \in (\frac{-3\pi }{2},\frac{-2\pi }{3})\cup (\frac{2\pi }{3},\frac{7\pi }{6})$
$\Rightarrow \varphi \in (\frac{2\pi }{3},\frac{7\pi }{6})$
Among the given options clearly $\frac{\pi }{2}<\varphi <\frac{4\pi }{3}$ is the only set contains all $\varphi$ values.