In the expansion of ${\left(\frac{\mathrm{x}}{\cos \mathrm{\theta }}+\frac{1}{\mathrm{xsin}\mathrm{\theta }}\right)}^{16}$ if l₁ is the least value of the term independent of x when $\mathrm{\pi }/8\le \mathrm{\theta }\le \mathrm{\pi }/4$ and l₂ is the least value of the term independent of x when $\mathrm{\pi }/16\le \mathrm{\theta }\le \mathrm{\pi }/8$, then the ratio ${\mathrm{l}}_2 : {\mathrm{l}}_1$ is

${\mathrm{T}}_{\mathrm{r}+1}{=}^{16}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{\mathrm{x}}{\cos \mathrm{\theta }}\right)}^{16-\mathrm{r}}{\left(\frac{1}{\mathrm{xsin}\mathrm{\theta }}\right)}^{\mathrm{r}}{=}^{16}{\mathrm{C}}_{\mathrm{r}}\frac{{\mathrm{x}}^{16-2\mathrm{r}}}{{\cos }^{16-\mathrm{r}}{\mathrm{\theta sin}}^{\mathrm{r}}\mathrm{\theta }}$. If $16-2\mathrm{r}=0$ then $\mathrm{r}=8$.

$\therefore$The term independent of x is ${\mathrm{T}}_9=\frac{{ }^{16}{\mathrm{C}}_8}{{\cos }^8{\mathrm{\theta sin}}^8\mathrm{\theta }}=\frac{{ }^{16}{\mathrm{C}}_8{2}^8}{{\sin }^{8}2\mathrm{\theta }}$

If $\frac{\mathrm{\pi }}{8}\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{4}$ the least value of ${\mathrm{T}}_9$ is ${\mathrm{l}}_1=\frac{{ }^{16}{\mathrm{C}}_8{2}^8}{{\sin }^8(\mathrm{\pi }/2)}{=}^{16}{\mathrm{C}}_8{2}^8$

If $\frac{\mathrm{\pi }}{16}\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{8}$ then the least value of ${\mathrm{T}}_9$ is ${\mathrm{l}}_2=\frac{{ }^{16}{\mathrm{C}}_8{2}^8}{{\sin }^8(\mathrm{\pi }/4)}{=}^{16}{\mathrm{C}}_8{2}^8\times 2^4$

${\mathrm{l}}_2:{\mathrm{l}}_1{=}^{16}{\mathrm{C}}_8{2}^8\times 2^4{:}^{16}{\mathrm{C}}_8{2}^8=16:1$