Q83 / 200 · Manipal (MET) 2010 · Mathematics
If $\cos x=\frac{2 \cos y-1}{2-\cos y}$, where $x, y \in(0, \pi)$, then $\tan \frac{x}{2} \cdot \cot \frac{y}{2}$ is equal to
- A$\sqrt{2}$
- Correct answerB$\sqrt{3}$
- C$\frac{1}{\sqrt{2}}$
- D$\frac{1}{\sqrt{3}}$
Solution
$\cos x=\frac{2 \cos y-1}{2-\cos y}$
$\Rightarrow \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}=\frac{2\left(\frac{1-\tan ^2 \frac{y}{2}}{1+\tan ^2 \frac{y}{2}}\right)-1}{2-\left(\frac{1-\tan ^2 \frac{y}{2}}{1+\tan ^2 \frac{y}{2}}\right)}$
$\Rightarrow \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}=\frac{2-2 \tan ^2 \frac{y}{2}-1-\tan ^2 \frac{y}{2}}{2+2 \tan ^2 \frac{y}{2}-1+\tan ^2 \frac{y}{2}}$
$\Rightarrow \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}=\frac{1-3 \tan ^2 \frac{y}{2}}{1+3 \tan ^2 \frac{y}{2}}$
$\Rightarrow 1+3 \tan ^2 \frac{y}{2}-\tan ^2 \frac{x}{2}-3 \tan ^2 \frac{x}{2} \tan ^2 \frac{y}{2}$
$=1-3 \tan ^2 \frac{y}{2}+\tan ^2 \frac{x}{2}-3 \tan ^2 \frac{x}{2} \tan ^2 \frac{y}{2}$
$\Rightarrow 6 \tan ^2 \frac{y}{2}=2 \tan ^2 \frac{x}{2}$
$\Rightarrow \tan \frac{x}{2} \cdot \cot \frac{y}{2}=\sqrt{3}$
Hence, the correct answer is option: 'B'.