6 In $\triangle A B C$, if $A \leqslant B \leqslant C$, then the range of $\sin A+\sin B-\sin C$ is ( )
A. $(0,1)$
B. $\left(0, \sqrt{3}-\frac{1}{2}\right]$
C. $\left(0, \frac{\sqrt{3}}{2}\right]$
D. $(0,2)$
6 C. From "the sum of any two sides of a triangle must be greater than the third side" and the Law of Sines, we know that $\sin A + \sin B - \sin C > 0$, and when $B = C$ and $A \rightarrow 0^\circ$, $\sin A + \sin B - \sin C \rightarrow 0$.
Given $A \leqslant B \leqslant C$, we know $C \geqslant \frac{\pi}{3}$, and since $A + B = \pi - C$, we have
$$
\begin{aligned}
\sin A + \sin B - \sin C & = 2 \cdot \sin \frac{A+B}{2} \cdot \cos \frac{A-B}{2} - \sin C \\
& \leqslant 2 \cdot \cos \frac{C}{2} \cdot 1 - \sin C \\
& = 2 \cdot \cos \frac{C}{2} \cdot \left(1 - \sin \frac{C}{2}\right).
\end{aligned}
$$

Since $\cos \frac{C}{2}$ and $1 - \sin \frac{C}{2}$ are both strictly decreasing positive functions, we have
$$
2 \cdot \cos \frac{C}{2} \cdot \left(1 - \sin \frac{C}{2}\right) \leqslant 2 \cdot \cos \frac{\pi}{6} \left(1 - \sin \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2},
$$

Therefore, $\sin A + \sin B - \sin C \leqslant \frac{\sqrt{3}}{2}$, and the equality holds when $A = B = C = \frac{\pi}{3}$.