Question

If $\cos x=\frac{2\cos y-1}{2-\cos y}$, where $x,y\in (0,\pi )$ and $\tan \frac{x}{2}\cot \frac{y}{2}=k$, then the value of $\left[k\right]$ is
(where [.] represents greatest integer function)

Answer 1

$\cos x=\frac{2\cos y-1}{2-\cos y}$
$\Rightarrow \frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{\frac{2(1-{\tan }^2\frac{y}{2})}{1+{\tan }^2\frac{y}{2}}-1}{2-\frac{1-{\tan }^2\frac{y}{2}}{1+{\tan }^2\frac{y}{2}}}\Rightarrow \frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{2-2{\tan }^2\frac{y}{2}-1-{\tan }^2\frac{y}{2}}{2+2{\tan }^2\frac{y}{2}-1+{\tan }^2\frac{y}{2}}\Rightarrow \frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{1-3{\tan }^2\frac{y}{2}}{1+3{\tan }^2\frac{y}{2}}$
Applying componendo and dividendo
$\Rightarrow \frac{2}{-2{\tan }^2\frac{x}{2}}=\frac{2}{-6{\tan }^2\frac{y}{2}}$
$\Rightarrow 3{\tan }^2\frac{y}{2}={\tan }^2\frac{x}{2}\Rightarrow {\tan }^2\frac{x}{2}{\cot }^2\frac{y}{2}=3\Rightarrow \tan \frac{x}{2}\cot \frac{y}{2}=\pm \sqrt{3}$
As $x,y\in (0,\pi )$, so
$\frac{x}{2},\frac{y}{2}\in (0,\frac{\pi }{2})$

$\therefore \tan \frac{x}{2}\cot \frac{y}{2}=\sqrt{3}=k$
$\Rightarrow \left[k\right]=1$