If cos-1cosα+cosβ1+cosα cosβ=2Tan-1x then x=

Let a=tan2 α2 , b=tan2 β2 cos-11-a1+a+1-b1+b1+1-a1+a·1-b1+b=cos-11+b-a-ab+1+a-b-ab1+a+b+ab+1-a-b+ab =cos-12-2ab2+2ab=cos-11-ab1+ab=2tan-1 ab =2tan-1tan α2 tan β2 ⇒ x=tan α2 tan β2

If cos-1cosα+cosβ1+cosα cosβ=2Tan-1x then x=

Let a=tan2 α2 , b=tan2 β2 cos-11-a1+a+1-b1+b1+1-a1+a·1-b1+b=cos-11+b-a-ab+1+a-b-ab1+a+b+ab+1-a-b+ab =cos-12-2ab2+2ab=cos-11-ab1+ab=2tan-1 ab =2tan-1tan α2 tan β2 ⇒ x=tan α2 tan β2

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If $\cos^{- 1} [\frac{cosα + cosβ}{1 + cosα cosβ}] = 2 {Tan}^{- 1} x$ then x=

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$\tan \frac{α}{2} \tan \frac{β}{2}$

$tanα tanβ$

$\cot \frac{α}{2} \cot \frac{β}{2}$

$\cot α \cot β$

answer is A.

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Detailed Solution

$L e t a = \tan^{2} \frac{α}{2}, b = \tan^{2} \frac{β}{2} \cos^{- 1} [\frac{\frac{1 - a}{1 + a} + \frac{1 - b}{1 + b}}{1 + \frac{1 - a}{1 + a} \cdot \frac{1 - b}{1 + b}}] = \cos^{- 1} [\frac{1 + b - a - ab + 1 + a - b - ab}{1 + a + b + ab + 1 - a - b + ab}] = \cos^{- 1} [\frac{2 - 2 ab}{2 + 2 ab}] = \cos^{- 1} (\frac{1 - ab}{1 + ab}) = 2 \tan^{- 1} (\sqrt{ab}) = 2 \tan^{- 1} (\tan \frac{α}{2} \tan \frac{β}{2}) \Rightarrow x = \tan \frac{α}{2} \tan \frac{β}{2}$