`sec^-1((1+tan ^2x)/(1-tan ^2 x))`

Answer

To solve the expression \( \sec^{-1}\left(\frac{1 + \tan^2 x}{1 - \tan^2 x}\right) \), we can follow these steps:

### Step-by-step Solution:

1. **Recall the Identity**:
We know that \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \). Therefore, we can rewrite the expression using this identity.

2. **Rewrite the Expression**:
The expression can be rewritten as:
\[ \frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1 + \frac{\sin^2 x}{\cos^2 x}}{1 - \frac{\sin^2 x}{\cos^2 x}} = \frac{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}}{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{1}{\cos^2 x - \sin^2 x} \]
since \( \sin^2 x + \cos^2 x = 1 \).

3. **Simplify the Denominator**:
We can simplify the denominator:
\[ \cos^2 x - \sin^2 x = \cos 2x \]
Thus, we have:
\[ \frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1}{\cos 2x} \]

4. **Substitute Back**:
Now substituting back into the original expression:
\[ \sec^{-1}\left(\frac{1}{\cos 2x}\right) = \sec^{-1}(\sec 2x) \]

5. **Final Result**:
Since \( \sec^{-1}(\sec \theta) = \theta \) for \( \theta \) in the range of \( \sec^{-1} \), we have:
\[ \sec^{-1}(\sec 2x) = 2x \]

Thus, the final answer is:
\[ \sec^{-1}\left(\frac{1 + \tan^2 x}{1 - \tan^2 x}\right) = 2x \]