period the function `f(x)=(sin{sin(n x)})/(tan (x/n))`, n`in` N, is `6pi` then `n=`--------

Options

A. 3
B. 2
C. 1
D. none of these

Explanation

To find the value of \( n \) such that the period of the function \( f(x) = \frac{\sin(\sin(nx))}{\tan(x/n)} \) is \( 6\pi \), we will analyze the periods of the numerator and denominator separately.

### Step 1: Determine the period of the numerator
The numerator is \( \sin(\sin(nx)) \). The function \( \sin(x) \) has a period of \( 2\pi \). Since \( \sin(nx) \) is a composition of the sine function, we need to find its period.

- The period of \( \sin(nx) \) is given by:
\[ \text{Period of } \sin(nx) = \frac{2\pi}{n} \]

- Therefore, the period of \( \sin(\sin(nx)) \) remains the same as \( \sin(nx) \) because the outer sine function does not change the period:
\[ T_1 = \frac{2\pi}{n} \]

### Step 2: Determine the period of the denominator
The denominator is \( \tan(x/n) \). The function \( \tan(x) \) has a period of \( \pi \). For \( \tan(x/n) \), the period is adjusted based on the coefficient of \( x \).

- The period of \( \tan(x/n) \) is given by:
\[ \text{Period of } \tan(x/n) = n\pi \]

### Step 3: Find the overall period of the function
The overall period of the function \( f(x) \) is the least common multiple (LCM) of the periods of the numerator and denominator.

- The periods we have are:
\[ T_1 = \frac{2\pi}{n}, \quad T_2 = n\pi \]

- The LCM of \( T_1 \) and \( T_2 \) can be calculated as:
\[ \text{LCM}\left(\frac{2\pi}{n}, n\pi\right) = \frac{\text{LCM}(2\pi, n^2\pi)}{\text{GCD}(n, 1)} \]

- Since \( \text{GCD}(n, 1) = 1 \), we have:
\[ \text{LCM}(2\pi, n^2\pi) = 2n\pi \]

### Step 4: Set the overall period equal to \( 6\pi \)
We know that the overall period is given as \( 6\pi \):
\[ 2n\pi = 6\pi \]

### Step 5: Solve for \( n \)
To find \( n \), we can simplify:
\[ 2n = 6 \]
\[ n = \frac{6}{2} = 3 \]

Thus, the value of \( n \) is \( 3 \).

### Final Answer
\[ n = 3 \]