Answer:
In \[\Delta \text{ }ABC\] \[\angle A+\angle B+\angle C=180{}^\circ \] \[\angle A+\angle C=180{}^\circ -\angle B\] Divide by 2 on both sides \[\frac{\angle A+\angle C}{2}=\frac{180{}^\circ -\angle B}{2}\] \[\frac{\angle A+\angle C}{2}=90{}^\circ -\frac{\angle B}{2}\] \[\sin \left( \frac{\angle A+\angle C}{2} \right)=\sin \left( 90{}^\circ -\frac{\angle B}{2} \right)\] \[\sin \left( \frac{\angle A+\angle C}{2} \right)=\cos \frac{\angle B}{2}\] \[\sin \frac{(A+c)}{2}=\cos \frac{B}{2}\] Hence Proved.
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