ntIn any triangle ABC, prove the following: asin[A/2].sin[(B-C)/2] + bsin[B/2].sin[(C-A)/2] + csin[C/2].sin[(A-B)/2] = 0n

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If : a+b+c=pie

then prove that : sin2A+sin2B+sin2C= 4sinAsinBsinC

\frac{a^{2} - c^{2}}{b^{2}} = \frac{\sin (A - C)}{\sin (A + C)}

\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C} = 0

\sin (\frac{B - C}{2}) = \frac{b - c}{a} \cos \frac{A}{2}

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