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 $(0,\frac{\pi }{2})$ is

Since $a=(\frac{4}{\sin x}+\frac{1}{1-\sin x})$, a is least.

$\therefore \frac{da}{dx}=[-\frac{4}{{\sin }^{2}x}+\frac{1}{(1-\sin x{)}^2}]\cos x=0$

We have to find the values of $x$  in the interval  $(0,\pi /2)$ . Thus, $\cos x\ne 0$  and the other factor when equated to zero gives $\sin x=2/3$. Now,

$\frac{d^{2}a}{d{x}^2}=[-\frac{4}{{\sin }^{2}x}+\frac{1}{(1-\sin x{)}^2}](-\sin x)$$+[\frac{8}{{\sin }^{3}x}+\frac{2}{(1-\sin x{)}^3}]{\cos }^{2}x$

Put  $\sin x=\frac{2}{3}$ and  ${\cos }^{2}x=1-\frac{4}{9}=\frac{5}{9}$

$\therefore \frac{d^{2}a}{d{x}^2}=0+[\frac{8}{8/27}+2\times 27]\frac{5}{9}=81\times \frac{5}{9}=45>0$

Thus, a is minimum and its value is  $\frac{4}{2/3}+\frac{1}{1-(2/3)}=6+3=9$