Question

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

• 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: (A), (C), (D)

$0<\phi <\frac{\pi }{2}$
$\frac{3\pi }{2}<\phi <2\pi$
$\frac{4\pi }{3}<\phi <\frac{3\pi }{2}$

$2cos\theta (1-sin\phi )=si{n}^2\theta (tan\frac{\theta }{2}+cot\frac{\theta }{2})cos\phi -1$

$\Rightarrow 2cos\theta (1-sin\phi )=si{n}^2\theta \left(\frac{2}{sin\theta }cos\phi \right)-1$

$\Rightarrow 2cos\theta -2cossin\phi =2sin\theta cos\phi -1$

$\Rightarrow 2cos\theta +1=2sin(\theta +\phi ).....\left(i\right)$

Also given that $tan(2\pi -\theta )>0$

$\Rightarrow tan\theta <0.....\left(1\right)$
$-1<sin\theta <-\frac{\sqrt{3}}{2}$

$\Rightarrow \theta ϵ(\frac{3\Pi }{2},\frac{5\Pi }{3}).....\left(2\right)$

So, ${}^{\prime }{\theta }^{\prime }$ is in 4th quadrant $\Rightarrow$ L.H.S. of equation (i) will be positive.

$1<2cos\theta +1<2$

$\Rightarrow 1<2sin(\theta +\phi )<2$

$\Rightarrow \frac{1}{2}<sin\left(\theta \phi \right)<1$

$\Rightarrow 2\pi +\frac{\pi }{6}<\theta +\phi <\frac{5\pi }{6}+2\pi$

$\Rightarrow 2\pi +\frac{\pi }{6}-{\theta }_{max}<\phi <2\pi +\frac{5\pi }{6}-{\theta }_{min}$

$\Rightarrow \frac{\pi }{2}<\phi <\frac{4\pi }{3}$