The value of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\), \(|x| < \frac{1}{2}, x \neq 0\)

Let \(\mathrm{T}=\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\)
Put \(x^2=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2\)
\(\begin{aligned}
\therefore \quad \mathrm{T} & =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right) \\
& =\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) \\
& =\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right) \\
& =\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\
& =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \\
& =\frac{\pi}{4}+\theta \\
& =\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2
\end{aligned}\)