Given, $x={\log }_e\left[\cot \left(\frac{\pi }{4}+\theta \right)\right]$

$\Rightarrow e^x=\cot \left(\frac{\pi }{4}+\theta \right)$

            $=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }$

Now, $\cos \text{h}x=\frac{e^x+e^{-x}}{2}$

             $=\left\{\frac{e^x+{\displaystyle \frac{1}{e^x}}}{2}\right\}$

            $=\frac{1}{2}\left\{\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }+\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right\}$

            $=\frac{1}{2}\left\{\frac{{\left(\cos \theta -\sin \theta \right)}^2+{\left(\cos \theta +\sin \theta \right)}^2}{\left(\cos \theta +\sin \theta \right)\left(\cos \theta -\sin \theta \right)}\right\}$

             $=\frac{1}{2}\left\{\frac{2}{\left({\cos }^2\theta -{\sin }^2\theta \right)}\right\}$

              $=\left\{\frac{1}{\cos 2\theta }\right\}$

               $=\sec 2\theta$

And, $\sin \text{h}x=\frac{e^x-e^{-x}}{2}$

            $=\left\{\frac{e^x-{\displaystyle \frac{1}{e^x}}}{2}\right\}$

             $=\frac{1}{2}\left\{\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }-\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right\}$

               $=\frac{1}{2}\left\{\frac{{\left(\cos \theta -\sin \theta \right)}^2-{\left(\cos \theta +\sin \theta \right)}^2}{\left(\cos \theta +\sin \theta \right)\left(\cos \theta -\sin \theta \right)}\right\}$

              $=\frac{1}{2}\left\{\frac{-4\cos \theta \sin \theta }{\left({\cos }^2\theta -{\sin }^2\theta \right)}\right\}$

              $=\left\{\frac{-2\cos \theta \sin \theta }{\cos 2\theta }\right\}$

 $=-\left\{\frac{\sin 2\theta }{\cos 2\theta }\right\}$

$=-\tan 2\theta$