The value of \(\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x\) is

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