In the expansion of  ${(\frac{x}{\cos \theta }+\frac{1}{x\sin \theta })}^{16},$  if  $l_1$  is the least value of the term independent of  x  when $\frac{\pi }{8}\le \theta \le \frac{\pi }{4}$ and $l_2$  is the least value of the term independent of x when $\frac{\pi }{16}\le \theta \le \frac{\pi }{8},$ then the ratio  $l_2:l_1$ is equal to :

$T_{r+1}={{16}_C}_{{ }_r}{\left(\frac{x}{\cos \theta }\right)}^{16-r}{\left(\frac{1}{x\sin \theta }\right)}^r$

$={16}_{C_r}{\left(x\right)}^{16-r}\times \frac{1}{{\left(\cos \theta \right)}^{16-r}{\left(\sin \theta \right)}^r}$

For independent of x:  $16-r=0\Rightarrow r=8$
 $$\begin{gathered} \Rightarrow T_9={16}_{C_r}\frac{1}{{\cos }^8\theta {\sin }^8\theta } \\ ={{16}_C}_{{ }_8}\frac{2^8}{{\left(\sin 2\theta \right)}^8} \\ for\theta \in [\frac{\pi }{8},\frac{\pi }{4}]l_{1}isforleastfor{\theta }_1=\frac{\pi }{4} \\ for\theta \in [\frac{\pi }{16},\frac{\pi }{8}]l_{2}isforleastfor{\theta }_1=\frac{\pi }{8} \\ \frac{l_2}{l_1}=\frac{{\left(\sin 2{\theta }_1\right)}^8}{{\left(\sin 2{\theta }_2\right)}^8}={\left(\sqrt{2}\right)}^8=\frac{16}{1} \end{gathered}$$