Let `f (x) =cos ^(-1) ((1-tan ^(2)(x//2))/(1+ tan ^(2) (x//2)))` . Solve for f(x).

Answer

To solve for \( f(x) = \cos^{-1} \left( \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \right) \), we can follow these steps:

### Step 1: Recognize the Identity
We start by recognizing the trigonometric identity:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
In our case, let \( \theta = \frac{x}{2} \). Therefore, we can write:
\[ \frac{1 - \tan^2 \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} = \cos(x) \]

### Step 2: Substitute into the Function
Now we substitute this identity back into our function:
\[ f(x) = \cos^{-1}(\cos(x)) \]

### Step 3: Apply the Inverse Cosine Property
Using the property of inverse trigonometric functions, we know that:
\[ \cos^{-1}(\cos(x)) = x \quad \text{(for } x \text{ in the range } [0, \pi] \text{)} \]
However, since \( \cos^{-1} \) is defined for angles in the range \( [0, \pi] \), we need to consider the periodic nature of the cosine function.

### Step 4: Final Result
Thus, we conclude:
\[ f(x) = x \quad \text{(for } x \text{ in the appropriate range)} \]

### Final Answer
Hence, the solution is:
\[ f(x) = x \]

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