The minimum value of $\alpha$, where $\alpha =\frac{4}{\sin x}+\frac{1}{1-\sin x}$, is 

 

Clearly, $f\left(x\right)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$Clearly, $f\left(x\right)$is definer for all $x\ne n\pi$ and $x\ne (4n+1)\frac{\pi }{2},n\in Z$

Now,

$f^{\mathrm{\prime }}\left(x\right)=-4\mathrm{cosec}x\cot x+\frac{\cos x}{(1-\sin x{)}^2}=\frac{-4\cos x}{{\sin }^{2}x}+\frac{\cos x}{(1-\sin x{)}^2}$

For $f\left(x\right)$to be maximum or minimum, we must have

$f^{\mathrm{\prime }}\left(x\right)=0$

$\Rightarrow \frac{-4\cos x}{{\sin }^{2}x}+\frac{\cos x}{(1-\sin x{)}^2}=0$

$\Rightarrow \frac{4}{{\sin }^{2}x}=\frac{1}{(1-\sin x{)}^2}\left[\because \cos x\ne 0\text{ for }x\ne (4n+1)\frac{\pi }{2}\right.$

$\begin{array}{l}\Rightarrow 3{\sin }^{2}x-8\sin x+4=0\\ \Rightarrow (3\sin x-2)(\sin x-2)=0\Rightarrow 3\sin x-2=0\Rightarrow \sin x=\frac{2}{3}\end{array}$

Now, $\begin{array}{r}{f}^{\prime \prime }\left(x\right)=4\mathrm{cosec}x{\cot }^{2}x+4{\mathrm{cosec}}^{3}x-\frac{\sin x}{(1-\sin x{)}^2}\\ +\frac{2{\cos }^{2}x}{(1-\sin x)}\end{array}$

$\Rightarrow f^{\prime \prime }\left(\frac{2}{3}\right)>0$

Thus,$f\left(x\right)$ is minimum when$\sin x=\frac{2}{3}$.

$\therefore \alpha =4\times \frac{3}{2}+\frac{1}{(1-2/3)}=6+3=9$