Discussion Forum

Let $\theta$ , $\phi \epsilon [0,2\pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin^{2} \theta$

$\left( \tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right)\cos \phi-1$,

 $\tan (2\pi-\theta)>0$

  and $-1 < \sin \theta < -\frac{-\sqrt{3}}{2}$ then , $\phi$ cannot satisfy 

Answer:

Explanation:

Concept lnvolved. It is based on range of sin x i.e., [-1,1] and the internal for a< x< b.

 Description  of situation As $\theta, \phi \epsilon [0,2\pi]$ and

  $\tan (2 \pi-\theta) 0,-1<\sin \theta <-\frac{\sqrt{3}}{2}$

 $\tan (2 \pi-\theta) >0$ $\Rightarrow$  $-\tan \theta >0$

 $\therefore$  $\theta \epsilon$ II or IV quadrant

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

 $\Rightarrow$   $\frac{4 \pi}{3} < \theta < \frac{5 \pi}{3}$ but $\theta \epsilon$ II or IV  quadrant 

 $\Rightarrow$    $\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$ .....(i)

 Sol. here

 $2 \cos \theta (1-\sin \phi )$

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

 $\Rightarrow$  $2 \cos \theta-2 \cos \theta \sin  \phi$

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

 =$2 \cos \theta-2 cos \theta \sin \phi$

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

 =$2 \cos \theta +1=2 \sin \phi \cos \theta+2 \sin \theta \cos \phi$

         $\Rightarrow$  $2 \cos \theta+1=2 \sin ( \theta+\phi)$.....................(ii)

 From Eq.(i)

   $\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$ $\Rightarrow 2 \cos \theta+1 \epsilon(1,2)$

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

$\rightarrow$   $\frac{1}{2} < \sin (\theta +\phi) <1$......(iii)

 $\Rightarrow$  $\frac{\pi}{6} < \theta +\phi < \frac{5 \pi}{6}$ or $\frac{13 \pi}{6} < \theta +\phi < \frac{17 \pi}{6}$

 $\therefore$  $\frac{\pi}{6} -\theta < \phi < \frac{5 \pi}{6}- \theta$

 or $\frac{13 \pi}{6}-\theta < \phi  < \left(\frac{17 \pi}{6}\right)-\theta$

 $\Rightarrow$ $\phi \epsilon\left(-\frac{3  \pi}{2},-\frac{2\pi}{3}\right)or \left(\frac{2\pi}{3},\frac{7 \pi}{6}\right),$

as $\theta \epsilon\left(\frac{3  \pi}{2},\frac{5 \pi}{3}\right)$