Differentiation of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. \(\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right)\), is

Let \(\mathrm{u}=\tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]\)
and \(v=\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right]\)
Put \(x=\tan \theta\), then \(\theta=\tan ^{-1} x\)
\(
\begin{aligned}
\therefore \mathrm{u} & =\tan ^{-1}\left[\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right]-\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right] \\
& =\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right]=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] \\
& =\tan ^{-1}\left[\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right]-\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}
\end{aligned}
\)
\(
\therefore \quad \mathrm{u}=\frac{\tan ^{-1} x}{2}
\)
\(
\begin{aligned}
v & =\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+\tan ^2 \theta}}{2 \sqrt{1+\tan ^2 \theta}}}\right] \\
& =\cos ^{-1}\left[\sqrt{\frac{1+\sec \theta}{2 \sec \theta}}\right] \\
& =\cos ^{-1}\left[\sqrt{\frac{1+\frac{1}{\cos \theta}}{\frac{2}{\cos \theta}}}\right] \\
& =\cos ^{-1}\left[\sqrt{\frac{1+\cos \theta}{2}}\right] \\
& =\cos ^{-1}\left(\sqrt{\frac{2 \cos ^2\left(\frac{\theta}{2}\right)}{2}}\right) \\
& =\cos ^{-1}\left(\cos \frac{\theta}{2}\right)-\frac{0}{2}
\end{aligned}
\)
\(
\therefore \quad v=\frac{\tan ^{-1} x}{2}
\)
From (i) and (ii), we get \(\mathrm{u}=\mathrm{v}\)
\(
\therefore \quad \frac{d u}{d v}=1
\)