If minimum value of term free from `x` for `(x/(sintheta) + 1/(xcostheta))^(16)` is `L_1` in `[pi/8,pi/4]` and `L_2` in `[pi/16,pi/8]`, then `L_2/L_1`

Answer

To solve the problem, we need to find the ratio \( \frac{L_2}{L_1} \) where \( L_1 \) and \( L_2 \) are the minimum values of the term free from \( x \) in the expression \( \left( \frac{x}{\sin \theta} + \frac{1}{x \cos \theta} \right)^{16} \) evaluated over specified intervals.

### Step-by-Step Solution

1. **Identify the General Term**:
The general term in the binomial expansion of \( \left( \frac{x}{\sin \theta} + \frac{1}{x \cos \theta} \right)^{16} \) is given by:
\[ T_r = \binom{16}{r} \left( \frac{x}{\sin \theta} \right)^{16-r} \left( \frac{1}{x \cos \theta} \right)^r \]
Simplifying this, we have:
\[ T_r = \binom{16}{r} \frac{x^{16-r}}{\sin^{16-r} \theta} \cdot \frac{1}{x^r \cos^r \theta} = \binom{16}{r} \frac{x^{16-2r}}{\sin^{16-r} \theta \cos^r \theta} \]

2. **Finding the Term Free from \( x \)**:
For the term to be free from \( x \), the exponent of \( x \) must be zero:
\[ 16 - 2r = 0 \implies r = 8 \]
Thus, the term free from \( x \) is:
\[ T_8 = \binom{16}{8} \frac{1}{\sin^8 \theta \cos^8 \theta} \]

3. **Simplifying the Expression**:
We can rewrite \( T_8 \) as:
\[ T_8 = \binom{16}{8} \frac{1}{(\sin \theta \cos \theta)^8} = \binom{16}{8} \frac{2^8}{\sin(2\theta)^8} \]
This is because \( \sin(2\theta) = 2 \sin \theta \cos \theta \).

4. **Calculating \( L_1 \)**:
We need to evaluate \( L_1 \) in the interval \( \left[\frac{\pi}{8}, \frac{\pi}{4}\right] \). Since \( \sin(2\theta) \) is increasing in this interval, we take the upper limit \( \theta = \frac{\pi}{4} \):
\[ L_1 = \binom{16}{8} \frac{2^8}{\sin(2 \cdot \frac{\pi}{4})^8} = \binom{16}{8} \frac{2^8}{1^8} = \binom{16}{8} \cdot 2^8 \]

5. **Calculating \( L_2 \)**:
Now, we evaluate \( L_2 \) in the interval \( \left[\frac{\pi}{16}, \frac{\pi}{8}\right] \). Again, since \( \sin(2\theta) \) is increasing, we take the upper limit \( \theta = \frac{\pi}{8} \):
\[ L_2 = \binom{16}{8} \frac{2^8}{\sin(2 \cdot \frac{\pi}{8})^8} = \binom{16}{8} \frac{2^8}{\left(\frac{1}{\sqrt{2}}\right)^8} = \binom{16}{8} \cdot 2^8 \cdot 2^4 = \binom{16}{8} \cdot 2^{12} \]

6. **Finding the Ratio \( \frac{L_2}{L_1} \)**:
Now we can find the ratio:
\[ \frac{L_2}{L_1} = \frac{\binom{16}{8} \cdot 2^{12}}{\binom{16}{8} \cdot 2^8} = \frac{2^{12}}{2^8} = 2^{4} = 16 \]

### Final Answer
Thus, the value of \( \frac{L_2}{L_1} \) is \( 16 \).