1708\hspace{-16}$Given $\bf{\frac{4}{\sin x}+\frac{1}{1-\sin x}=k}$\\\\\\ Using $\bf{Cauchy-Schwarcz}$ Inequality::\\\\\\ $\bf{\frac{a^2}{x}+\frac{b^2}{y}\geq \frac{(a+b)^2}{x+y}}$\\\\\\ Equality Hold When $\bf{\frac{a^2}{x} = \frac{b^2}{y}}$\\\\\\ So $\bf{\frac{2^2}{\sin x}+\frac{1^2}{1-\sin x}\geq \frac{(2+1)^2}{\sin x+1-\sin x}}$\\\\\\ So $\bf{k\geq 9}$ and equality hold when $\bf{\frac{2^2}{\sin x} = \frac{1^2}{1-\sin x}}$\\\\\\ So $\bf{k_{Min.}=9}$ when $\bf{\sin x = \frac{4}{5}}$\\\\\\ where $\bf{x=\sin^{-1}\left(\frac{4}{5}\right)\in \left(0,\frac{\pi}{2}\right)}$
