$\cot^{- 1} (\sqrt{\cos α}) - {Tan}^{- 1} (\sqrt{\cos α}) = x \geq 0, then \sin x =$

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$\tan^{2} \frac{α}{2}$

$\cot \frac{α}{2}$

$\tan α$

$\cot^{2} (\frac{α}{2})$

answer is A.

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Detailed Solution

$\begin{array}{l} \cot^{- 1} (\sqrt{\cos α}) - \tan^{- 1} (\sqrt{\cos α}) = x \\ \Rightarrow \frac{π}{2} - 2 \tan^{- 1} (\sqrt{\cos α}) = x \\ \Rightarrow \sin x = \sin (\frac{π}{2} - 2 \tan^{- 1} (\sqrt{\cos α})) \end{array}$
$put \tan^{- 1} (\sqrt{\cos α}) = θ$
$\begin{array}{l} \Rightarrow \tan θ = \sqrt{\cos α} \\ \sin x = \sin (\frac{π}{2} - 2 θ) = \cos 2 θ \\ = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} \\ = \frac{1 - \cos α}{1 + \cos α} \\ \sin x = \frac{1 - \cos α}{1 + \cos^{} α} \\ = \frac{2 \sin^{2} α / 2}{2 \cos^{2} α / 2} \\ \Rightarrow \sin x = \tan^{2} \frac{α}{2} \end{array}$