If $\cot (\mathrm{\theta }-\mathrm{\alpha }), 3\cot \mathrm{\theta }, \cot (\mathrm{\theta }+\mathrm{\alpha })$are in AP (where, $\mathrm{\theta }\ne \frac{\mathrm{n\pi }}{2},\mathrm{\alpha }\ne \mathrm{k\pi },\mathrm{k}\in \mathrm{I}$) then $\frac{2{\sin }^2\mathrm{\theta }}{{\sin }^2\mathrm{\alpha }}$is equal to ______

we have $6\cot \mathrm{\theta }=\cot (\mathrm{\theta }-\mathrm{\alpha })+\cot (\mathrm{\theta }+\mathrm{\alpha })$
$\begin{array}{l}\Rightarrow \frac{6\cos \mathrm{\theta }}{\sin \mathrm{\theta }}=\frac{\sin 2\mathrm{\theta }}{\sin (\mathrm{\theta }-\mathrm{\alpha })+\sin (\mathrm{\theta }+\mathrm{\alpha })}\\ \Rightarrow \frac{6\cos \mathrm{\theta }}{\sin \mathrm{\theta }}=\frac{2\sin \mathrm{\theta cos}\mathrm{\theta }}{{\sin }^2\mathrm{\theta }-{\sin }^2\mathrm{\alpha }}\\ \Rightarrow 3\left({\sin }^2\mathrm{\theta }-{\sin }^2\mathrm{\alpha }\right)={\sin }^2\mathrm{\theta }\\ \text{ or }2{\sin }^2\mathrm{\theta }=3{\sin }^2\mathrm{\alpha }\Rightarrow \frac{2{\sin }^2\mathrm{\theta }}{{\sin }^2\mathrm{\alpha }}=3\end{array}$