Let `f(x)=cos^(-1)((1-tan^(2)(x)/(2))/(1+tan^(2)(x)/(2)))`. Then which of the following statement is/are true ?

Options

A. Ranges of f(x) is `[0,pi)`
B. `f(x)=pi` has infinite roots
C. y = f(x) is identical with `y = cos^(-1)(cos x)`
D. y = f(x) has period `2pi`

Explanation

To solve the problem, we need to analyze the function given by:

\[ f(x) = \cos^{-1} \left( \frac{1 - \tan^2(x)/2}{1 + \tan^2(x)/2} \right) \]

### Step 1: Simplify the expression inside the inverse cosine function

We can use the trigonometric identity:

\[ \frac{1 - \tan^2(x)}{1 + \tan^2(x)} = \cos(2x) \]

In our case, we have:

\[ \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} = \cos(x) \]

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \cos^{-1}(\cos(x)) \]

### Step 2: Determine the value of \( f(x) \)

The function \( \cos^{-1}(\cos(x)) \) gives us \( x \) in the range of \( [0, \pi] \). Therefore, we can express \( f(x) \) as:

\[ f(x) = x \quad \text{for } x \in [0, \pi] \]

### Step 3: Identify the range of \( f(x) \)

Since \( f(x) = x \) for \( x \in [0, \pi] \), the range of \( f(x) \) is:

\[ \text{Range of } f(x) = [0, \pi] \]

### Step 4: Check periodicity of \( f(x) \)

The function \( f(x) \) is not periodic in the traditional sense because it is defined as \( f(x) = x \) within the range \( [0, \pi] \). However, if we consider the periodic nature of the cosine function, we can say that:

\[ f(x + 2\pi) = f(x) \]

This means that the function exhibits periodic behavior with a period of \( 2\pi \).

### Step 5: Conclusion about the statements

From the analysis, we can conclude:

1. The range of \( f(x) \) is \( [0, \pi] \).
2. The function has a period of \( 2\pi \).

### Final Answer

Thus, the true statements regarding \( f(x) \) are:

- The range of \( f(x) \) is \( [0, \pi] \).
- \( f(x) \) has a period of \( 2\pi \).