If x=logcotπ4+θ then sinhx=___

x=log cotπ4+θ⇒ex=cotπ4+θe−x=tanπ4+θsinhx=ex−e−x2=cotπ4+θ−tanπ4+θ2=2cotπ2+2θ2=−tan2θ

If x=logcotπ4+θ then sinhx=___

x=log cotπ4+θ⇒ex=cotπ4+θe−x=tanπ4+θsinhx=ex−e−x2=cotπ4+θ−tanπ4+θ2=2cotπ2+2θ2=−tan2θ

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If $x = \log [\cot (\frac{π}{4} + θ)]$ then $\sinh x =___$

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$\tan 2 θ$

$\cot 2 θ$

$- \tan 2 θ$

$- \cot 2 θ$

answer is C.

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Detailed Solution

$x = \log c o t (\frac{π}{4} + θ) \Rightarrow e^{x} = c o t (\frac{π}{4} + θ)$
$e^{- x} = \tan (\frac{π}{4} + θ)$
$\sinh x = \frac{e^{x} - e^{- x}}{2} = \frac{c o t (\frac{π}{4} + θ) - \tan (\frac{π}{4} + θ)}{2}$
$\begin{array}{l} = \frac{2 c o t (\frac{π}{2} + 2 θ)}{2} \\ = - \tan 2 θ \end{array}$