If $\cos^{- 1} [\frac{\cos α + \cos β}{1 + \cos α . \cos β}] = 2 \tan^{- 1} x$ then $' x' =$

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\tan \frac{α}{2} . \tan \frac{β}{2}

\tan α . \tan β

\cot \frac{α}{2} . \cot \frac{β}{2}

\cot α . \cot β

answer is A.

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Detailed Solution

\cos^{- 1} [\frac{\cos α + \cos β}{1 + \cos α . \cos β}] = 2 \tan^{- 1} x

\Rightarrow \cos^{- 1} [\frac{\cos α + \cos β}{1 + \cos α . \cos β}] = \cos^{- 1} [\frac{1 - x^{2}}{1 + x^{2}}]

$\frac{\cos α + \cos β}{1 + \cos α . \cos β} = \frac{1 - x^{2}}{1 + x^{2}} \Rightarrow \frac{\cos α + \cos β + 1 + \cos α . \cos β}{\cos α + \cos β - 1 - \cos α . \cos β} = \frac{1 - x^{2} + 1 + x^{2}}{1 - x^{2} - 1 - x^{2}}$

$\frac{(\cos α + 1) + \cos β (1 + \cos α)}{\cos α - 1 + \cos β (1 - \cos α)} = \frac{2}{- 2 x^{2}}$

$\frac{(1 + \cos α) (1 + \cos β)}{(1 - \cos α) (- 1 + \cos β)} = - \frac{1}{x^{2}} \Rightarrow \frac{(1 + \cos α) (1 + \cos β)}{(1 - \cos α) (1 - \cos β)} = \frac{1}{x^{2}}$

$\frac{2 \cos^{2} \frac{α}{2} .2 \cos^{2} \frac{β}{2}}{2 \sin^{2} \frac{α}{2} .2 \sin^{2} \frac{β}{2}} = \frac{1}{x^{2}} \Rightarrow x^{2} = \frac{\sin^{2} \frac{α}{2} . \sin^{2} \frac{β}{2}}{\cos^{2} \frac{α}{2} . \cos^{2} \frac{β}{2}}$