in triangle ABC ,show that sinA+sinB -sinC= 4 sin(A/2) sin(B2)sin(C2)

Dear Student,
Recheck your R.H.S. part.

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{sinA}+\mathrm{sinB}-\mathrm{sinC} \\ =\mathrm{sinA}+\left(\mathrm{sinB}-\mathrm{sinC}\right) \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\cos \left(\frac{\mathrm{A}}{2}\right)+2\cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)\sin \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{B}+\mathrm{C}}{2}\right)+2\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{A}}{2}\right)\sin \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)+2\sin \left(\frac{\mathrm{A}}{2}\right)\sin \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\left[\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)+\sin \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\right] \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\left[2\sin \left(\frac{\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)+\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{2}\right)\cos \left(\frac{\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)-\left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}{2}\right)\right] \\ =2\sin \left(\frac{\mathrm{A}}{2}\right)\left[2\sin \left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{C}}{2}\right)\right] \\ =4\sin \left(\frac{\mathrm{A}}{2}\right)\sin \left(\frac{\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{C}}{2}\right) \\ =\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence} \mathrm{Proved}. \end{gathered}$$

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