If $\cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) - \cos^{- 1} (\frac{1 - b^{2}}{1 + b^{2}}) = 2 \tan^{- 1} x$ , then ‘x’ is equal to

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$\frac{a - b}{1 + a b}$

$\frac{b - a}{1 + b a}$

$\frac{a + b}{1 - a b}$

None of these

answer is A.

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Detailed Solution

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

$∴ 2 \tan^{- 1} a - 2 \tan^{- 1} b = 2 \tan^{- 1} x$

$\Rightarrow \tan^{- 1} a - \tan^{- 1} b = \tan^{- 1} x$

$\Rightarrow \tan^{- 1} (\frac{a - b}{1 + a b}) = \tan^{- 1} x$

$∴ x = \frac{a - b}{1 + a b}$

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