$\cot \theta=\frac{1+\cos 2 \theta}{\sin 2 \theta}=\{ \therefore 1+\cos 2 \theta=2 \cos ^{2} \theta \,\& \, \sin 2 \theta=2 \sin \theta \cos \theta\}$

put, $\theta=\frac{\pi}{24}$

$\left\{\therefore \cos \frac{\pi}{12}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \, \& \, \sin \frac{\pi}{12}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\right\}$

$\Rightarrow \cot \left(\frac{\pi}{24}\right)=\frac{1+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)}$

$=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$

$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$