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Q.

If $x = \log [\cot (π / 4 + θ)]$ then $\cos h x =$

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a

$\sec 2 θ$

b

$\cos e c 2 θ$

c

$\cot 2 θ$

d

$\tan 2 θ$

answer is A.

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Detailed Solution

$x = \log [\cot (\frac{π}{4} + θ)] \Rightarrow e^{x} = \frac{\cot θ - 1}{\cot θ + 1}$

$\cosh x = \frac{e^{x} + e^{- x}}{2} = \frac{1 + \tan^{2} θ}{1 - \tan^{2} θ} = \sec 2 θ$

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