Prove that `tan^(-1)[(sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))]=pi/4+1/2cos^(-1)x^2`

Explanation

To prove that

\[ \tan^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2) \]

we will follow these steps:

### Step 1: Substitute \( x^2 = \cos(2\theta) \)

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

\[ \sqrt{1+x^2} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2}\cos(\theta) \]

\[ \sqrt{1-x^2} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2}\sin(\theta) \]

### Step 2: Substitute these into the original expression

Now substitute these values into the left-hand side:

\[ \tan^{-1}\left(\frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{\sqrt{2}\cos(\theta) - \sqrt{2}\sin(\theta)}\right) \]

This simplifies to:

\[ \tan^{-1}\left(\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)}\right) \]

### Step 3: Use the tangent addition formula

Using the tangent addition formula, we know that:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \]

In our case, we can express \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Therefore,

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \frac{\sin(\theta)}{\cos(\theta)}}{1 - \frac{\sin(\theta)}{\cos(\theta)}} = \frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)} \]

Thus,

\[ \tan^{-1}\left(\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)}\right) = \frac{\pi}{4} + \theta \]

### Step 4: Substitute back for \( \theta \)

Since \( \theta = \frac{1}{2} \cos^{-1}(x^2) \), we can substitute back:

\[ \tan^{-1}\left(\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)}\right) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2) \]

### Conclusion

Thus, we have shown that:

\[ \tan^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2) \]

Hence, the proof is complete.

---