$\cos^{- 1} (\frac{3 + 5 \cos x}{5 + 3 \cos x}) =$

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2 \tan^{- 1} (\frac{1}{2} \tan \frac{x}{2})

\frac{1}{2} \tan^{- 1} (2 \tan \frac{x}{2})

2 \tan^{- 1} (2 \tan \frac{x}{2})

\tan^{- 1} (\frac{1}{2} \tan \frac{x}{2})

answer is D.

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Detailed Solution

Let

\cos^{- 1} (\frac{3 + 5 \cos x}{5 + 3 \cos x}) = θ

$\Rightarrow \cos θ = \frac{3 + 5 \cos x}{5 + 3 \cos x}$

$\Rightarrow \frac{1 + \cos θ}{1 - \cos θ} = \frac{5 + 3 \cos x + 3 + 5 \cos x}{5 + 3 \cos x - 3 - 5 \cos x}$

$\Rightarrow \frac{1 + \cos θ}{1 - \cos θ} = \frac{8 (1 + \cos x)}{2 (1 - \cos x)}$

$\Rightarrow \cot^{2} \frac{θ}{2} = 4 \cot^{2} \frac{x}{2}$

$\Rightarrow \tan^{2} \frac{θ}{2} = \frac{1}{4} \tan^{2} \frac{x}{2}$

$\Rightarrow \tan \frac{θ}{2} = \frac{1}{2} \tan \frac{x}{2}$

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

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