Define: `cos^(-1)((1-x^(2))/(1+x^(2)))` in terms of `tan^(-1)x`

Explanation

To express \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) in terms of \( \tan^{-1}x \), we can follow these steps:

### Step 1: Set up the equation
Let:
\[ \theta = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]
This implies:
\[ \cos \theta = \frac{1-x^2}{1+x^2} \]

### Step 2: Use the identity for cosine
We know that:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \]
We can represent this in a right triangle where the adjacent side is \( 1 - x^2 \) and the hypotenuse is \( 1 + x^2 \).

### Step 3: Find the opposite side
Using the Pythagorean theorem:
\[ \text{opposite}^2 + (1 - x^2)^2 = (1 + x^2)^2 \]
Expanding both sides:
\[ \text{opposite}^2 + (1 - 2x^2 + x^4) = (1 + 2x^2 + x^4) \]
Simplifying:
\[ \text{opposite}^2 + 1 - 2x^2 + x^4 = 1 + 2x^2 + x^4 \]
This leads to:
\[ \text{opposite}^2 = 4x^2 \]
Thus:
\[ \text{opposite} = 2\sqrt{x} \]

### Step 4: Find \( \tan \theta \)
Now we can find \( \tan \theta \):
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2\sqrt{x}}{1 - x^2} \]

### Step 5: Express \( \theta \) in terms of \( \tan^{-1} \)
Thus, we can express \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{x}}{1 - x^2}\right) \]

### Final Result
Therefore, we have:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2\sqrt{x}}{1 - x^2}\right) \]