Let `f (x) =cos ^(-1) ((1-tan ^(2)(x//2))/(1+ tan ^(2) (x//2)))`
Which of the following statement (s) is/are correct about `f (x)` ?

Options

A. Domain is R
B. Range is `[0,pi]`
C. `f (x)` is even
D. `f (x)` is dervable in `(pi, 2pi)`

Answer

To analyze the function \( f(x) = \cos^{-1} \left( \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \right) \), we will check the correctness of the statements regarding its domain, range, evenness, and differentiability.

### Step 1: Simplifying the Function
We start with the expression inside the inverse cosine function:
\[ f(x) = \cos^{-1} \left( \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \right) \]
Using the identity for cosine of double angles, we know that:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Setting \(\theta = \frac{x}{2}\), we can rewrite:
\[ f(x) = \cos^{-1}(\cos(x)) = x \quad \text{for } x \in [0, \pi] \]

### Step 2: Checking the Domain
The function \( f(x) \) is defined for all real numbers since the argument of the inverse cosine function can take any value between -1 and 1, which corresponds to the range of \(\cos\).

**Conclusion**: The domain of \( f(x) \) is \( \mathbb{R} \).

### Step 3: Checking the Range
The range of \( \cos^{-1}(x) \) is always \( [0, \pi] \). Therefore, since \( f(x) \) simplifies to \( x \) for \( x \in [0, \pi] \), we can conclude that the range of \( f(x) \) is also \( [0, \pi] \).

**Conclusion**: The range of \( f(x) \) is \( [0, \pi] \).

### Step 4: Checking if \( f(x) \) is Even
A function \( f(x) \) is even if \( f(-x) = f(x) \). We can check:
\[ f(-x) = \cos^{-1} \left( \frac{1 - \tan^2\left(-\frac{x}{2}\right)}{1 + \tan^2\left(-\frac{x}{2}\right)} \right) = \cos^{-1} \left( \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \right) = f(x) \]
Thus, \( f(x) \) is an even function.

**Conclusion**: \( f(x) \) is even.

### Step 5: Checking Differentiability
The function \( f(x) \) is derived from the cosine function, which is differentiable everywhere. Therefore, \( f(x) \) is differentiable in the interval \( (0, 2\pi) \).

**Conclusion**: \( f(x) \) is differentiable in \( (0, 2\pi) \).

### Final Conclusion
All the statements regarding \( f(x) \) are correct:
1. Domain is \( \mathbb{R} \) - True
2. Range is \( [0, \pi] \) - True
3. \( f(x) \) is even - True
4. \( f(x) \) is differentiable in \( (0, 2\pi) \) - True