=cos-1a-ba+b =cos-11-ba1+ba tan2θ=ba then θ=tan-1ba =cos-11-tan2θ1+tan2θ =2θ =2tan-1ba

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${Cos}^{- 1} (\frac{a - b}{a + b}) =$

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$2 \tan^{- 1} b$

$2 \tan^{- 1} \sqrt{\frac{b}{a}}$

$2 \tan^{- 1} \sqrt{\frac{a}{b}}$

$2 \tan^{- 1} (\frac{ab}{a - b})$

answer is A.

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Detailed Solution

$= \cos^{- 1} (\frac{a - b}{a + b}) = \cos^{- 1} (\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}) \tan^{2} θ = \frac{b}{a} t h e n θ = \tan^{- 1} (\sqrt{\frac{b}{a}}) = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) = 2 θ = 2 \tan^{- 1} \sqrt{\frac{b}{a}}$