Example 9 Simplify: $\arctan \frac{1+|x|-\sqrt{1-x^{2}}}{1+|x|+\sqrt{1-x^{2}}}+\frac{1}{2} \arccos |x|(-1 \leqslant x \leqslant 1)$.
From the problem, we have \( |x| \in [0,1], \frac{1+|x|-\sqrt{1-x^{2}}}{1+|x|+\sqrt{1-x^{2}}}=\theta \), then \( 2 \theta+\varphi \in [0, \pi] \).
Let \( z=\left(1+|x|+\sqrt{1-x^{2}}\right)+\left(1+|x|-\sqrt{1-x^{2}}\right) \mathrm{i}, \omega=|x|+\sqrt{1-x^{2}} \mathrm{i} \), then \( \arg \left(z^{2} \cdot \omega\right)=2 \theta+\varphi \).
It is easy to get \( z^{2} \cdot \omega=4(1+|x|) \mathrm{i} \), so \( \arg \left(z^{2} \cdot \omega\right)=\frac{\pi}{2} \). Therefore, the original expression \( =\theta+\frac{\varphi}{2}=\frac{2 \theta+\varphi}{2}=\frac{\pi}{4} \).