Different w.r.t tan-1 x when x ≠ 0.

Let $y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $z=\tan ^{-1} x$

Now, put $x=\tan \theta$

$\therefore y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)$ and $z=\tan ^{-1}(\tan \theta)=\theta$

So, $\quad \tan \left(\frac{\sqrt{\sec \theta}-1}{\tan }\right)=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$

$\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$

$\tan ^{-1}\left(\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}\right)=\tan ^{-1}\left(\frac{\sin \theta / 2}{\cos \theta / 2}\right)$

$\Rightarrow y=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \Rightarrow y=\frac{\theta}{2}$

Differentiating both parametric functions w.r.t. $\theta$

$\frac{d y}{d \theta}=\frac{1}{2} \cdot \frac{d}{d \theta}(\theta) \quad$ and $\quad \frac{d z}{d \theta}=\frac{d}{d \theta}(\theta)$

$=\frac{1}{2} \cdot 1=\frac{1}{2} \quad$ and $\quad \frac{d z}{d \theta}=1$

Thus, $\frac{d y}{d z}=\frac{d y / d \theta}{d z / d \theta}=\frac{1 / 2}{1}=\frac{1}{2}$.