If `tan^(-1)( (sqrt(1+x^2)-sqrt(1-x^2))/(sqrt((1+x^2))+sqrt((1-x^2))))`

Options

A. `cos x`
B. `(pi)/4-(1/2)cos^(-1)(x^2)`
C. `tan x`
D. none

Answer

To solve the expression \( \tan^{-1} \left( \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \right) \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Substitution**:
Let \( x^2 = \cos(2\theta) \). Then, we can express \( \sqrt{1+x^2} \) and \( \sqrt{1-x^2} \) in terms of \( \theta \):
\[ \sqrt{1+x^2} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2} \cos(\theta) \]
\[ \sqrt{1-x^2} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2} \sin(\theta) \]

**Hint**: Use trigonometric identities to simplify the square roots.

2. **Substituting into the expression**:
Substitute these values back into the original expression:
\[ \tan^{-1} \left( \frac{\sqrt{2} \cos(\theta) - \sqrt{2} \sin(\theta)}{\sqrt{2} \cos(\theta) + \sqrt{2} \sin(\theta)} \right) \]

3. **Factor out \( \sqrt{2} \)**:
\[ = \tan^{-1} \left( \frac{\sqrt{2} (\cos(\theta) - \sin(\theta))}{\sqrt{2} (\cos(\theta) + \sin(\theta))} \right) \]
\[ = \tan^{-1} \left( \frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)} \right) \]

**Hint**: Simplifying fractions can often reveal useful identities.

4. **Using the tangent subtraction formula**:
Recognize that:
\[ \frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)} = \tan\left(\frac{\pi}{4} - \theta\right) \]
Therefore, we can rewrite the expression:
\[ = \tan^{-1} \left( \tan\left(\frac{\pi}{4} - \theta\right) \right) \]

**Hint**: Recall that \( \tan^{-1}(\tan(x)) = x \) when \( x \) is in the principal range of the arctangent function.

5. **Final expression**:
Thus, we have:
\[ = \frac{\pi}{4} - \theta \]

6. **Finding \( \theta \)**:
Since \( x^2 = \cos(2\theta) \), we can find \( \theta \):
\[ 2\theta = \cos^{-1}(x^2) \implies \theta = \frac{1}{2} \cos^{-1}(x^2) \]

7. **Substituting back**:
Therefore, we substitute \( \theta \) back into our expression:
\[ = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}(x^2) \]

### Conclusion:
The final answer is:
\[ \tan^{-1} \left( \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}(x^2) \]