The least value of' a' for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$ has at least one solution in the interval $(0,\pi /2),$ is

The equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$

will have at least one solution in the interval $(0,\pi /2)$ if $a$ is the minimum value of $f\left(x\right)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in the interval 

$(0,\pi /2).$

Now,

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

For maximum and minimum, we must have 

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=0\\ \Rightarrow \frac{-4\cos x}{{\sin }^{2}x}+\frac{\cos x}{(1-\sin x{)}^2}=0\\ \Rightarrow \cos x\left\{-\frac{4}{{\sin }^{2}x}+\frac{1}{(1-\sin x{)}^2}\right\}=0\\ \Rightarrow \frac{4}{{\sin }^{2}x}=\frac{1}{(1-\sin x{)}^2}\\ \Rightarrow 2(1-\sin x)=\sin x\Rightarrow \sin x=\frac{2}{3}\Rightarrow x={\sin }^{-1}\left(\frac{2}{3}\right)\end{array}$

It can be easily checked that $f^{\prime \prime }\left(x\right)>0$ for $\sin x=\frac{2}{3}$

The minimum value of $f\left(x\right)$ is given by

$f\left({\sin }^{-1}\frac{2}{3}\right)=\frac{4}{2/3}+\frac{1}{1-2/3}=6+3=9.$

Hence, $a=9.$