2tan−1⁡tan⁡θ2tan⁡ϕ2=

We have, 2tan−1⁡tan⁡θ2tan⁡ϕ2=cos−1⁡1−tan2⁡θ2tan2⁡ϕ21+tan2⁡θ2tan2⁡ϕ2=cos−1⁡1-sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ21+sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ2=cos−1⁡cos2⁡θ2cos2⁡ϕ2−sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ2+sin2⁡θ2sin2⁡ϕ2=cos−1cos⁡θ2cos⁡ϕ2+sin⁡θ2sin⁡ϕ2cos⁡θ2cos⁡ϕ2−sin⁡θ2sin⁡ϕ2cos⁡θ2cos⁡ϕ2+sin⁡θ2sin⁡ϕ22−2sin⁡θ2cos⁡θ2sin⁡ϕ2cos⁡ϕ2=cos−1⁡cos⁡θ−ϕ2cos⁡θ+ϕ2cos2⁡θ−ϕ2−sin⁡θsin⁡ϕ2=cos−1⁡2cos⁡θ−ϕ2cos⁡θ+ϕ22cos2⁡θ−ϕ2−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡(θ−ϕ)−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ+sin⁡θsin⁡ϕ−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ

2tan−1⁡tan⁡θ2tan⁡ϕ2=

We have, 2tan−1⁡tan⁡θ2tan⁡ϕ2=cos−1⁡1−tan2⁡θ2tan2⁡ϕ21+tan2⁡θ2tan2⁡ϕ2=cos−1⁡1-sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ21+sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ2=cos−1⁡cos2⁡θ2cos2⁡ϕ2−sin2⁡θ2sin2⁡ϕ2cos2⁡θ2cos2⁡ϕ2+sin2⁡θ2sin2⁡ϕ2=cos−1cos⁡θ2cos⁡ϕ2+sin⁡θ2sin⁡ϕ2cos⁡θ2cos⁡ϕ2−sin⁡θ2sin⁡ϕ2cos⁡θ2cos⁡ϕ2+sin⁡θ2sin⁡ϕ22−2sin⁡θ2cos⁡θ2sin⁡ϕ2cos⁡ϕ2=cos−1⁡cos⁡θ−ϕ2cos⁡θ+ϕ2cos2⁡θ−ϕ2−sin⁡θsin⁡ϕ2=cos−1⁡2cos⁡θ−ϕ2cos⁡θ+ϕ22cos2⁡θ−ϕ2−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡(θ−ϕ)−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ+sin⁡θsin⁡ϕ−sin⁡θsin⁡ϕ=cos−1⁡cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ

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$2 \tan^{- 1} (\tan \frac{θ}{2} \tan \frac{ϕ}{2}) =$

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$\cos^{- 1} (\frac{\cos θ + \cos ϕ}{1 + \cos θcos ϕ})$

$\cos^{- 1} (\frac{\cos θ - \cos ϕ}{1 + \cos θcos ϕ})$

$\cos^{- 1} (\frac{\cos θ + \cos ϕ}{1 - \cos θcos ϕ})$

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answer is A.

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Detailed Solution

We have, $2 \tan^{- 1} (\tan \frac{θ}{2} \tan \frac{ϕ}{2})$

$= \cos^{- 1} [\frac{1 - \tan^{2} \frac{θ}{2} \tan^{2} \frac{ϕ}{2}}{1 + \tan^{2} \frac{θ}{2} \tan^{2} \frac{ϕ}{2}}] = \cos^{- 1} [\frac{1 - \frac{\sin^{2} \frac{θ}{2} \sin^{2} \frac{ϕ}{2}}{\cos^{2} \frac{θ}{2} \cos^{2} \frac{ϕ}{2}}}{1 + \frac{\sin^{2} \frac{θ}{2} \sin^{2} \frac{ϕ}{2}}{\cos^{2} \frac{θ}{2} \cos^{2} \frac{ϕ}{2}}}]$

$\begin{array}{l} = \cos^{- 1} [\frac{\cos^{2} \frac{θ}{2} \cos^{2} \frac{ϕ}{2} - \sin^{2} \frac{θ}{2} \sin^{2} \frac{ϕ}{2}}{\cos^{2} \frac{θ}{2} \cos^{2} \frac{ϕ}{2} + \sin^{2} \frac{θ}{2} \sin^{2} \frac{ϕ}{2}}] \\ = \cos^{- 1} [\frac{(\cos \frac{θ}{2} \cos \frac{ϕ}{2} + \sin \frac{θ}{2} \sin \frac{ϕ}{2}) (\cos \frac{θ}{2} \cos \frac{ϕ}{2} - \sin \frac{θ}{2} \sin \frac{ϕ}{2})}{{(\cos \frac{θ}{2} \cos \frac{ϕ}{2} + \sin \frac{θ}{2} \sin \frac{ϕ}{2})}^{2} - 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \sin \frac{ϕ}{2} \cos \frac{ϕ}{2}}] \\ \begin{array}{l} = \cos^{- 1} [\frac{\cos (\frac{θ - ϕ}{2}) \cos (\frac{θ + ϕ}{2})}{\cos^{2} \frac{θ - ϕ}{2} - \frac{\sin θsin ϕ}{2}}] \\ = \cos^{- 1} [\frac{2 \cos (\frac{θ - ϕ}{2}) \cos (\frac{θ + ϕ}{2})}{2 \cos^{2} \frac{θ - ϕ}{2} - \sin θsin ϕ}] \\ = \cos^{- 1} [\frac{\cos θ + \cos ϕ}{1 + \cos (θ - ϕ) - \sin θsin ϕ}] \end{array} \\ \begin{array}{l} = \cos^{- 1} [\frac{\cos θ + \cos ϕ}{1 + \cos θcos ϕ + \sin θsin ϕ - \sin θsin ϕ}] \\ = \cos^{- 1} [\frac{\cos θ + \cos ϕ}{1 + \cos θcos ϕ}] \end{array} \end{array}$