In the expansion of `((x)/(costheta)+(1)/(xsintheta))^(16)`, if `l_(1)` is the least value of the term independent of `x` when `(pi)/(8) le theta le (pi)/(4)` and `l_(2)` is the least value of the term independent of `x` when `(pi)/(16) le theta le (pi)/(8)`, then the value of `(l_(2))/(l_(1))` is

Options

A. `8`
B. `32`
C. `16`
D. `64`

Explanation

To solve the problem, we need to find the least values of the term independent of \( x \) in the expansion of

\[ \left( \frac{x}{\cos \theta} + \frac{1}{x \sin \theta} \right)^{16} \]

for two different ranges of \( \theta \).

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the expansion can be expressed as:

\[ T_{r+1} = \binom{16}{r} \left( \frac{x}{\cos \theta} \right)^{16-r} \left( \frac{1}{x \sin \theta} \right)^{r} \]

This simplifies to:

\[ T_{r+1} = \binom{16}{r} \frac{x^{16-r}}{\cos^{16-r} \theta} \cdot \frac{1}{x^r \sin^r \theta} = \binom{16}{r} \frac{x^{16-2r}}{\cos^{16-r} \theta \sin^r \theta} \]

### Step 2: Determine the Condition for Independence from \( x \)
The term \( T_{r+1} \) will be independent of \( x \) when the exponent of \( x \) is zero:

\[ 16 - 2r = 0 \implies r = 8 \]

### Step 3: Find the Independent Term
Substituting \( r = 8 \) into the general term gives:

\[ T_{9} = \binom{16}{8} \frac{1}{\cos^8 \theta \sin^8 \theta} \]

### Step 4: Simplify the Independent Term
We can rewrite this as:

\[ T_{9} = \binom{16}{8} \frac{1}{\sin^8(2\theta)} \cdot 2^8 \]

### Step 5: Calculate \( L_1 \) for \( \frac{\pi}{8} \leq \theta \leq \frac{\pi}{4} \)
To find \( L_1 \), we need to minimize \( \sin^8(2\theta) \) over the interval \( \frac{\pi}{8} \leq \theta \leq \frac{\pi}{4} \):

- At \( \theta = \frac{\pi}{4} \), \( 2\theta = \frac{\pi}{2} \) and \( \sin(2\theta) = 1 \).
- At \( \theta = \frac{\pi}{8} \), \( 2\theta = \frac{\pi}{4} \) and \( \sin(2\theta) = \frac{\sqrt{2}}{2} \).

Thus, the minimum occurs at \( \theta = \frac{\pi}{4} \):

\[ L_1 = \binom{16}{8} \cdot 2^8 \]

### Step 6: Calculate \( L_2 \) for \( \frac{\pi}{16} \leq \theta \leq \frac{\pi}{8} \)
Now, we find \( L_2 \) for the interval \( \frac{\pi}{16} \leq \theta \leq \frac{\pi}{8} \):

- At \( \theta = \frac{\pi}{8} \), \( 2\theta = \frac{\pi}{4} \) and \( \sin(2\theta) = \frac{\sqrt{2}}{2} \).
- At \( \theta = \frac{\pi}{16} \), \( 2\theta = \frac{\pi}{8} \) and \( \sin(2\theta) = \sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} \).

To find the minimum, we evaluate \( \sin^8(2\theta) \):

The minimum occurs at \( \theta = \frac{\pi}{8} \):

\[ L_2 = \binom{16}{8} \cdot 2^8 \cdot \left( \frac{\sqrt{2}}{2} \right)^8 = \binom{16}{8} \cdot 2^8 \cdot \frac{1}{16} = \frac{\binom{16}{8} \cdot 2^8}{16} \]

### Step 7: Calculate \( \frac{L_2}{L_1} \)
Now we find the ratio:

\[ \frac{L_2}{L_1} = \frac{\frac{\binom{16}{8} \cdot 2^8}{16}}{\binom{16}{8} \cdot 2^8} = \frac{1}{16} \]

### Final Answer
Thus, the value of \( \frac{L_2}{L_1} \) is:

\[ \boxed{16} \]