A geometry problem by Rakhi Bhattacharyya

$\begin{aligned} (1-\cot 22^\circ)(1-\cot 23^\circ) & = \left(1 - \frac {\cos 22^\circ}{\sin 22^\circ}\right) \left(1 - \frac {\cos 23^\circ}{\sin 23^\circ} \right) \\ & = \frac {(\sin 22^\circ -\cos 22^\circ)(\sin 23^\circ -\cos 23^\circ)}{\sin 22^\circ \sin 23^\circ} \\ & = \frac {\sin 22^\circ \sin 23^\circ + \cos 22^\circ \cos 23^\circ - \sin 22^\circ\cos 23^\circ -\sin 23^\circ\cos 22^\circ }{\frac 12 \left(\cos (23^\circ - 22^\circ) - \cos (23^\circ + 22^\circ) \right)} \\ & = \frac {\cos 1^\circ - \sin 45^\circ}{\frac 12 \left(\cos 1^\circ - \cos 45^\circ \right)} \\ & = \frac {\cos 1^\circ - \frac 1{\sqrt 2}}{\frac 12 \left(\cos 1^\circ - \frac 1{\sqrt 2} \right)} \\ & = \boxed{2} \end{aligned}$

$\cot(45) = 1 \\ \cot(22 + 23) = 1 \\ \dfrac{(\cot 22 \cdot \cot 23 - 1)}{\cot 22 + \cot 23} = 1 \\ \cot 22 \cot 23 - 1= \cot 22 + \cot 23 \\ \cot 22 \cot 23 - \cot 22 - \cot 23 = 1 \\ \cot 22 \left(\cot 23 - 1\right) - \cot 23 = 1 \\ \cot 22 \left(\cot 23 - 1\right) - \cot 23 + 1 = 2 \\ \cot 22 \left(\cot 23 - 1\right) - \left(\cot 23 - 1\right) = 2 \\ (1-\cot 22°)(1-\cot 23°) = 2$