If $A+B+C={180}^o$, then $\sin A+\sin B-\sin C=$

The correct option is C $4\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}$
We have,

$\sin A+\sin B-\sin C$

$=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}-2\sin \frac{C}{2}\cos \frac{C}{2}$

Since, $\frac{A+B}{2}={90}^0-\frac{C}{2}$

$\sin \frac{A+B}{2}=\cos \frac{C}{2}$

Therefore,

$2\cos \frac{C}{2}\cdot \cos \frac{A-B}{2}-2\sin \frac{C}{2}\cdot \cos \frac{C}{2}2\cos \frac{C}{2}\left[\cos \frac{A-B}{2}-\sin \frac{C}{2}\right]\Rightarrow 2\cos \frac{C}{2}\left[\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right]\Rightarrow 2\cos \frac{C}{2}\left[2\sin \frac{A}{2}\cdot \sin \frac{B}{2}\right]\Rightarrow 4\sin \frac{A}{2}\cdot \sin \frac{B}{2}\cos \frac{C}{2}$

Hence, this is the answer.