Express in terms of : `"tan"^(-1)(2x)/(1-x^(2)` to `tan^(-1)x` for `xgt1`

Explanation

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

### Step 1: Use the double angle formula for tangent
We know that:
\[ \tan(2y) = \frac{2\tan(y)}{1 - \tan^2(y)} \]
This means:
\[ \tan^{-1}\left(\frac{2\tan(y)}{1 - \tan^2(y)}\right) = 2y \]

### Step 2: Substitute \(x\) in terms of \(\tan(y)\)
Let \(x = \tan(y)\). Then, we can write:
\[ \tan^{-1}\left(\frac{2\tan(y)}{1 - \tan^2(y)}\right) = 2y \]

### Step 3: Express \(y\) in terms of \(x\)
Since \(x = \tan(y)\), we have:
\[ y = \tan^{-1}(x) \]

### Step 4: Substitute \(y\) back into the equation
Now substituting \(y\) back, we get:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) \]

### Step 5: Adjust for the range \(x > 1\)
Since \(x > 1\), we need to adjust our result:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) - \pi \]

### Final Result
Thus, we can express:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) - \pi \quad \text{for } x > 1 \]

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