Explanation

(b)
Given, $x = \log\left\lbrack \cot\ \left( \frac{\pi}{4} + \theta \right) \right\rbrack$
$\Rightarrow \ \ \ e^{x} = \left\lbrack \cot\left( \frac{\pi}{4} + \theta \right) \right\rbrack$ …(i)
And $e^{- x} = \frac{1}{\cot\left( \frac{\pi}{4} + \theta \right)} = \tan\left( \frac{\pi}{4} + \theta \right)$ …(ii)
Now, $\sinh{x = \frac{e^{x} - e^{- x}}{2}}$
$= \frac{\cot{\left( \frac{\pi}{4} + \theta \right) - \tan\left( \frac{\pi}{4} + \theta \right)}}{2}$
$= \frac{1 - \tan^{2}\left( \frac{\pi}{4} + \theta \right)}{2\tan\left( \frac{\pi}{4} + \theta \right)} = \frac{1}{\tan{2\left( \frac{\pi}{4} + \theta \right)}}$
$= - \frac{1}{\cot{2\theta}} = - \tan{2\theta}$