To get \(\cot \frac{\pi}{24}\), we proceed as follows
\(\begin{aligned}
\cot \frac{\pi}{24} & =\frac{2 \cos ^2 \frac{\pi}{24}}{2 \sin \frac{\pi}{24} \cdot \cos \frac{\pi}{24}} \\
& =\frac{1+\cos \frac{\pi}{12}}{\sin \frac{\pi}{12}}=\frac{1+\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)}{1+\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)}
\end{aligned}\)
\(\begin{aligned}
& =\frac{1+\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}}{\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}} \\
& =2+\sqrt{2}+\sqrt{3}+\sqrt{6} \\
& \text {and } \tan \frac{\pi}{24}=-2+\sqrt{2}-\sqrt{3}+\sqrt{6} \\
& \text {So, } \frac{1}{2}\left(\tan \frac{\pi}{24}+\cot \frac{\pi}{24}\right)=\sqrt{2}+\sqrt{6}=\sqrt{2^2+2}+\sqrt{2} \\
& =\sqrt{a^2+a}+\sqrt{a} \quad \text { (given) } \\
\end{aligned}\)
So, \(a=2\)