- Home
- Standard 11
- Mathematics
मानाकि $\theta, \phi \in[0,2 \pi]$ इस प्रकार है कि $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta) > 0$ और $-1 < \sin \theta<-\frac{\sqrt{3}}{2}$. तब $\phi$ निम्न में से किसको संतुष्ट नहीं कर सकता ?
$(A)$ $0<\phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2}<\phi<\frac{4 \pi}{3}$
$(C)$ $\frac{4 \pi}{3}<\phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2}<\phi<2 \pi$
$(A,B,C)$
$(A,B,D)$
$(A,C,D)$
$(B,C,D)$
Solution
As $\tan (2 \pi-\theta)>0,-1<\sin \theta<-\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]$
$\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3}$
Now $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta(\tan \theta / 2+\cot \theta / 2) \cos \phi-1$
$\Rightarrow 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1$
$\Rightarrow 2 \cos \theta+1=2 \sin (\theta+\phi)$
As $\theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) \Rightarrow 2 \cos \theta+1 \in(1,2)$
$\Rightarrow 1<2 \sin (\theta+\phi)<2$
$\Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1$
As $\theta+\phi \in[0,4 \pi]$
$\Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ or $\theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right)$
$\Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta$ or $\frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta$
$\Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right)$
$\left(\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right)$
