If  $\cot (\theta -\alpha ),3\cot \theta ,\cot (\theta +\alpha )$  are in A.P. and  $\theta$  is not an integral multiple of  $\frac{\pi }{2},$ then the value 

of  $\frac{4{\sin }^2\theta }{3{\sin }^2\alpha }=$_____

Given  $\cot (\theta -\alpha ),3\cot \theta ,\cot (\theta +\alpha )$  are in A.P.
 $$\begin{gathered} \Rightarrow 6\cot \theta =\cot (\theta -\alpha )+\cot (\theta +\alpha ) \\ \Rightarrow \frac{6\cos \theta }{\sin \theta }=\frac{\sin 2\theta }{\sin (\theta +\alpha )\sin (\theta -\alpha )} \\ \Rightarrow 6\cos \theta \{{\sin }^2\theta -{\sin }^2\alpha \}=2{\sin }^2\theta \cos \theta \Rightarrow 3({\sin }^2\theta -{\sin }^2\alpha )={\sin }^2\theta \\ \Rightarrow 2{\sin }^2\theta =3{\sin }^2\alpha \\ \therefore \frac{2{\sin }^2\theta }{3{\sin }^2\alpha }=1 \end{gathered}$$