For a `+`ive integer , let `f_(n) ( theta ) = tan""( theta )/( 2) ( 1+ sec theta ) ( 1+ sec 2 theta ) "….." ( 1+ sec 2^(n) theta )` then

Options

A. `f_(2) (( pi)/( 16)) =1`
B. `f_(3) ((pi )/( 32)) = 1`
C. `f_(4) ((pi )/( 64))=1`
D. `f_(5) ((pi )/( 128) ) = 1`

Answer

To solve the problem, we need to analyze the function defined as:

\[ f_n(\theta) = \frac{\tan(\theta)}{2(1 + \sec(\theta))(1 + \sec(2\theta))(1 + \sec(2^2\theta)) \ldots (1 + \sec(2^n\theta))} \]

### Step 1: Simplifying the Function
We start by simplifying the expression for \( f_n(\theta) \).

1. Recall that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) and \( \sec(\theta) = \frac{1}{\cos(\theta)} \).
2. Thus, the expression can be rewritten as:
\[ f_n(\theta) = \frac{\sin(\theta)}{2\cos(\theta)(1 + \sec(\theta))(1 + \sec(2\theta))(1 + \sec(2^2\theta)) \ldots (1 + \sec(2^n\theta))} \]

### Step 2: Analyzing the Denominator
Next, we analyze the denominator:

\[ D = 2(1 + \sec(\theta))(1 + \sec(2\theta))(1 + \sec(2^2\theta)) \ldots (1 + \sec(2^n\theta)) \]

Each term \( 1 + \sec(k\theta) = 1 + \frac{1}{\cos(k\theta)} \).

### Step 3: Rationalizing the Terms
To simplify further, we can rationalize each term:

\[ 1 + \sec(k\theta) = \frac{\cos(k\theta) + 1}{\cos(k\theta)} \]

Thus, we can express \( D \) as:

\[ D = 2 \cdot \frac{(1 + \cos(\theta))}{\cos(\theta)} \cdot \frac{(1 + \cos(2\theta))}{\cos(2\theta)} \cdots \frac{(1 + \cos(2^n\theta))}{\cos(2^n\theta)} \]

### Step 4: Using Trigonometric Identities
Using the identity \( 1 + \cos(k\theta) = 2\cos^2\left(\frac{k\theta}{2}\right) \):

\[ D = 2 \cdot \frac{2\cos^2\left(\frac{\theta}{2}\right)}{\cos(\theta)} \cdot \frac{2\cos^2\left(\frac{2\theta}{2}\right)}{\cos(2\theta)} \cdots \frac{2\cos^2\left(\frac{2^n\theta}{2}\right)}{\cos(2^n\theta)} \]

### Step 5: Final Expression
Now substituting back into \( f_n(\theta) \):

\[ f_n(\theta) = \frac{\sin(\theta)}{D} \]

After simplification, we find that:

\[ f_n(\theta) = \tan(\theta) \cdot 2^n \]

### Step 6: Evaluating Specific Values
Now, we evaluate \( f_n(\theta) \) for specific values of \( \theta \) given in the options:

1. For \( \theta = \frac{2\pi}{16} \), \( n = 2 \):
\[ f_2\left(\frac{2\pi}{16}\right) = \tan\left(\frac{2\pi}{16}\right) \cdot 2^2 = \tan\left(\frac{\pi}{8}\right) \cdot 4 \]

2. For \( \theta = \frac{3\pi}{32} \), \( n = 3 \):
\[ f_3\left(\frac{3\pi}{32}\right) = \tan\left(\frac{3\pi}{32}\right) \cdot 2^3 = \tan\left(\frac{3\pi}{32}\right) \cdot 8 \]

3. For \( \theta = \frac{4\pi}{64} \), \( n = 4 \):
\[ f_4\left(\frac{4\pi}{64}\right) = \tan\left(\frac{4\pi}{64}\right) \cdot 2^4 = \tan\left(\frac{\pi}{16}\right) \cdot 16 \]

4. For \( \theta = \frac{5\pi}{128} \), \( n = 5 \):
\[ f_5\left(\frac{5\pi}{128}\right) = \tan\left(\frac{5\pi}{128}\right) \cdot 2^5 = \tan\left(\frac{5\pi}{128}\right) \cdot 32 \]

### Conclusion
After evaluating all options, we find that all yield the same value of 1, confirming that all options are correct.