If cos-1cos α+cos β1+cos α cos β=2Tan-1x then x=

cos-1cos α+cos β1+cos α cos β=2Tan-1x cosα+cosβ1+cosα cosβ=cos(2tan-1x) cosα+cosβ1+cosα cosβ=1-tan2(tan-1x)1+tan2(tan-1x) cosα+cosβ1+cosα cosβ=1-x21+x2 1+cosα cosβcosα+cosβ=1+x21-x2 1+cosα cosβ-cosα-cosβ1+cosα cosβ+cosα+cosβ=1+x2-1+x21+x2+1-x2 (1-cosα)(1-cosβ)(1+cosα)(1+cosβ)=x2 x=1-cosα1+cosα1-cosβ1+cosβ x=tanα2 tanβ2

If cos-1cos α+cos β1+cos α cos β=2Tan-1x then x=

cos-1cos α+cos β1+cos α cos β=2Tan-1x cosα+cosβ1+cosα cosβ=cos(2tan-1x) cosα+cosβ1+cosα cosβ=1-tan2(tan-1x)1+tan2(tan-1x) cosα+cosβ1+cosα cosβ=1-x21+x2 1+cosα cosβcosα+cosβ=1+x21-x2 1+cosα cosβ-cosα-cosβ1+cosα cosβ+cosα+cosβ=1+x2-1+x21+x2+1-x2 (1-cosα)(1-cosβ)(1+cosα)(1+cosβ)=x2 x=1-cosα1+cosα1-cosβ1+cosβ x=tanα2 tanβ2

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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

$\cos^{- 1} [\frac{\cos α + \cos β}{1 + \cos α \cos β}] = 2 {Tan}^{- 1} x \frac{cosα + cosβ}{1 + cosα cosβ} = \cos (2 \tan^{- 1} x) \frac{cosα + cosβ}{1 + cosα cosβ} = \frac{1 - \tan^{2} (\tan^{- 1} x)}{1 + \tan^{2} (\tan^{- 1} x)} \frac{cosα + cosβ}{1 + cosα cosβ} = \frac{1 - x^{2}}{1 + x^{2}} \frac{1 + cosα cosβ}{cosα + cosβ} = \frac{1 + x^{2}}{1 - x^{2}} \frac{1 + cosα cosβ - cosα - cosβ}{1 + cosα cosβ + cosα + cosβ} = \frac{1 + x^{2} - 1 + x^{2}}{1 + x^{2} + 1 - x^{2}} \frac{(1 - cosα) (1 - cosβ)}{(1 + cosα) (1 + cosβ)} = x^{2} x = \sqrt{\frac{1 - cosα}{1 + cosα}} \sqrt{\frac{1 - cosβ}{1 + cosβ}} x = \tan \frac{α}{2} \tan \frac{β}{2}$