Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \times$…
Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \times$ $\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0$ and
$-1 < \sin \theta < -\frac{\sqrt{3}}{2}$, then $\varphi$ cannot satisfy
- $0 < \varphi < \frac{\pi}{2}$
- $\frac{\pi}{2} < \varphi < \frac{4 \pi}{3}$
- $\frac{4 \pi}{3} < \varphi < \frac{3 \pi}{2}$
- $\frac{3 \pi}{2} < \varphi < 2 \pi$
Solution
As $\tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]$
Hence $\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$
Now $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1$
$\Rightarrow 2 \cos \theta(1-\sin \varphi)=2 \sin \theta \cos \varphi-1$
$\Rightarrow 2 \cos \theta+1=2 \sin (\theta+\varphi)$
As $\quad \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right), 1 < 2 \sin (\theta+\varphi) < 2$
As $\theta+\varphi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ or $(\theta+\varphi) \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right)$
We have $\varphi \in\left(-\frac{3 \pi}{2},-\frac{2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right)$
Asked in: JEE Advanced 2012 (Paper 1)
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