If A + B + C = π, prove that

Solution:

$=\sin (B+C-A)+\sin (C+A-B)-\sin (A+B-C)$

Using,

$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$=2 \sin C \cos (B-A)-\sin (A+B-C)$

Since $A+B+C=\pi$

$\rightarrow \mathrm{B}+\mathrm{A}=180-\mathrm{C}$

$=2 \sin C \cos (B-A)-\sin (\pi-C-C)$

$=2 \sin C \cos (B-A)-\sin 2 C$

Since , sin2A = 2sinAcosA,

$=2 \sin C \cos (B-A)-2 \sin C \cos C$

$=2 \sin C\{\cos (B-A)-\cos C\}$

Using ,

$\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)$

$=2 \sin C\left\{2 \sin \left(\frac{B-A+C}{2}\right) \sin \left(\frac{C-B+A}{2}\right)\right\}$

$=2 \sin C\left\{2 \sin \left(\frac{\pi-A-A}{2}\right) \sin \left(\frac{\pi-B-B}{2}\right)\right\}$

= 4cosAcosBsinC

= R.H.S