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Q)

If $\cos x=\large\frac{2\cos y-1}{2-\cos y}$ where $x,y\in (0,\pi)$ then $\tan\large\frac{x}{2}$$.\cot\large\frac{y}{2}$ is equal to

$(a)\;\sqrt 2\qquad(b)\;\sqrt 3\qquad(c)\;\large\frac{1}{\sqrt 2}$$\qquad(d)\;\large\frac{1}{\sqrt 3}$

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A)
$\cos x=\large\frac{2\cos y-1}{2-\cos y}$
$\qquad=\large\frac{1-\tan^2\Large\frac{x}{2}}{1+\tan^2\Large\frac{x}{2}}$
$\qquad=\large\frac{\Large\frac{2(1-\tan^2y/2)}{1+\tan^2y/2}\normalsize -1}{2-\Large\frac{1-\tan^2y/2}{1+\tan^2y/2}}$
$\qquad=6\tan^2\large\frac{y}{2}=$$2\tan^2\large\frac{x}{2}$
$\Rightarrow \tan^2\large\frac{x}{2}$$.\cot^2\large\frac{y}{2}$$=3$
$\Rightarrow \tan\large\frac{x}{2}$$\cot\large\frac{y}{2}$$=\sqrt 3$
Hence (b) is the correct answer.
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