If $x = \log_{e} [\cot (\frac{π}{4} + θ)]$ then $\tanh x =$

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$- \tan 2 θ$

$\tan 2 θ$

$- \cot 2 θ$

$\cos 2 θ$

answer is C.

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Detailed Solution

$G i v e n x = \log_{e} (c o t (\frac{π}{4} + θ)) \Rightarrow e^{x} = c o t (\frac{π}{4} + θ) = \frac{c o t θ - 1}{c o t θ + 1} \Rightarrow e^{x} = \frac{1 - \tan θ}{1 + \tan θ} a n d e^{- x} = \frac{1 + \tan θ}{1 - \tan θ} N o w e^{x} - e^{- x} = \frac{1 - \tan θ}{1 + \tan θ} - \frac{1 + \tan θ}{1 - \tan θ} = \frac{{(1 - \tan θ)}^{2} - {(1 + \tan θ)}^{2}}{(1 + \tan θ) (1 - \tan θ)} = \frac{- 4 \tan θ}{1 - \tan^{2} θ} s i m i l a r l y e^{x} + e^{- x} = \frac{2 (1 + \tan^{2} θ)}{1 - \tan^{2} θ} N o w \tan h x = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = \frac{- 2 \tan θ}{(1 + \tan^{2} θ)} = - \tan 2 θ$