Solution

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

Let \(x^2 = \cos 2\theta = \cos^2\theta – \sin^2\theta = 1 – 2\sin^2\theta = 2\cos^2\theta – 1\). Therefore, \[ \cos^{-1}x^2 = 2\theta \implies \theta = \frac{1}{2}\cos^{-1}x^2 \]

So, \[ \sqrt{1+x^2} = \sqrt{1 + \cos 2\theta} = \sqrt{2}\cos\theta \] \[ \sqrt{1-x^2} = \sqrt{1 – \cos 2\theta} = \sqrt{2}\sin\theta \]

Substituting back, \[ \text{LHS} = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta – \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\frac{\cos\theta + \sin\theta}{\cos\theta – \sin\theta}\right) \]

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

\[ = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2 = \text{RHS} \]

Hence proved.