If $4\mathrm{n\alpha }=\mathrm{\pi },$ then $\cot \mathrm{\alpha cot}2\mathrm{\alpha cot}3\mathrm{\alpha }\dots \cot (2\mathrm{n}-1)\mathrm{\alpha }$ is equal is

given $4n\alpha =\pi \Rightarrow 2n\alpha =\frac{\pi }{2}$
$\begin{array}{c}\mathrm{now},\cot \mathrm{\alpha }\cdot \cot (2\mathrm{n}-1)\mathrm{\alpha }=\cot \mathrm{\alpha cot}\left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)\\ =\cot \mathrm{\alpha }\cdot \tan \mathrm{\alpha }=1\end{array}$
similarly $\cot 2\alpha \cos (2n-2)\alpha =1$
$\cot 3\alpha \cdot \cot (2n-3)\alpha =1,\dots ,\cot (n-1)\alpha \cot (n+1)\alpha =1$
Thus, $\cot \alpha \cot 2\alpha \cot 3\alpha \dots \cot (2n-1)\alpha$
$=\{\cot \mathrm{\alpha cot}(2\mathrm{n}-1\left)\mathrm{\alpha }\right\}\{\cot 2\mathrm{\alpha cot}(2\mathrm{n}-2\left)\mathrm{\alpha }\right\}\dots \{\cot (\mathrm{n}-1)\mathrm{\alpha cot}(\mathrm{n}+1\left)\mathrm{\alpha }\right\}\cdot \cot \mathrm{n\alpha }$
$=1\cdot 1\cdot 1\dots 1\cdot 1\left[\because \cot n\alpha =\cot \frac{\pi }{4}=1\right]$
=1