G1.4Given that $0 \leq x_{0} \leq \frac{\pi}{2}$ and $x_{0}$ satisfies the equation $\sqrt{\sin x+1}-\sqrt{1-\sin x}=\sin \frac{x}{2}$. If $d=\tan x_{0}$, find the value of $d$.
$\begin{array}{l}(\sqrt{\sin x+1}-\sqrt{1-\sin x})^{2}=\left(\sin \frac{x}{2}\right)^{2} \\ 1+\sin x+1-\sin x-2 \sqrt{1-\sin ^{2} x}=\frac{1-\cos x}{2} \\ 2(2-2 \cos x)=1-\cos x \\ \cos x=1 \\ x_{0}=0 \\ d=\tan x_{0}=0\end{array}$