Given  $0\le \mathrm{x}\le \frac{1}{2}$ then the value of  $\tan \left[{\sin }^{-1} \left\{\frac{\mathrm{x}}{\sqrt{2}}+\frac{\sqrt{1-{\mathrm{x}}^2}}{\sqrt{2}}\right\}-{\sin }^{-1} \mathrm{x}\right]$ is

$$\begin{gathered} \begin{array}{r}\tan \left[{\sin }^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right\}-{\sin }^{-1}x\right]\\ =\tan \left[{\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}-{\sin }^{-1}x\right] \\ \end{array} \end{gathered}$$
$\text{ Put }{\sin }^{-1}x=\theta \Rightarrow x=\sin \theta$

$\begin{array}{l}=\tan \left[\left[{\sin }^{-1}\left\{\frac{\sin \theta +\sqrt{1-{\sin }^2\theta }}{\sqrt{2}}\right\}-\theta \right]\right.\\ =\tan \left[{\sin }^{-1}\left\{\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta \right\}-\theta \right]\\ =\tan \left[{\sin }^{-1}\left\{\sin \left(\theta +\frac{\pi }{4}\right)\right\}-\theta \right]\\ =\tan \left[\theta +\frac{\pi }{4}-\theta \right]\\ =\tan \frac{\pi }{4}=1\end{array}$