#### In the expansion of $\left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}$, if    is the least value of the term independent of $x$ when $\frac{\pi }{8}\leq \theta \leq \frac{\pi }{4}$ and    is the least value of the term independent of $x$ when $\frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}$, then the ratio   is equal to :  Option: 1 Option: 2 Option: 3 Option: 4

General Term of Binomial Expansion$\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}$

$\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}$

Term independent of x: It means term containing x0,

Now,

\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}

$\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16$

Correct option 1