Prove that:

$\frac{\pi}{8}=\left(22 \frac{1}{2}\right)^{\circ}$

Let $A=\left(22 \frac{1}{2}\right)^{\circ}$

Using the identity $\cot 2 A=\frac{\cot ^{2} A-1}{2 \cot A}$, we get

$\cot 45^{\circ}=\frac{\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-1}{2 \cot \left(22 \frac{1}{2}\right)^{\circ}}$

$\Rightarrow 1=\frac{\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-1}{2 \cot \left(22 \frac{1}{2}\right)^{\circ}} \quad\left(\because \cot 45^{\circ}=1\right)$

$\Rightarrow 2 \cot \left(22 \frac{1}{2}\right)^{\circ}-\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}+1=0$

$\Rightarrow \cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-2 \cot \left(22 \frac{1}{2}\right)^{\circ}-1=0$

$\Rightarrow\left\{\cot ^{2}\left(22 \frac{1}{2}\right)^{\circ}-2 \cot \left(22 \frac{1}{2}\right)^{\circ}+1\right\}-2=0$

$\Rightarrow\left\{\cot \left(22 \frac{1}{2}\right)^{\circ}-1\right\}^{2}=2$

$\Rightarrow \cot \left(22 \frac{1}{2}\right)^{\circ}-1=\sqrt{2}$

$\Rightarrow \cot \left(22 \frac{1}{2}\right)^{\circ}=\sqrt{2}+1$