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प्रश्न
In ∆ABC, prove that: \[a \sin\frac{A}{2} \sin \left( \frac{B - C}{2} \right) + b \sin \frac{B}{2} \sin \left( \frac{C - A}{2} \right) + c \sin \frac{C}{2} \sin \left( \frac{A - B}{2} \right) = 0\]
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उत्तर
Consider
\[a\sin\frac{A}{2}\sin\left( \frac{B - C}{2} \right) + b\sin\frac{B}{2}\sin\left( \frac{C - A}{2} \right) + c\sin\frac{C}{2}\sin\left( \frac{A - B}{2} \right)\]
\[ = k\left[ \sin\left\{ \pi - \left( B + C \right) \right\}\sin\frac{A}{2}\sin\left( \frac{B - C}{2} \right) + \sin\left\{ \pi - \left( C + A \right) \right\} \sin\frac{B}{2}\sin\left( \frac{C - A}{2} \right) + \sin\left\{ \pi - \left( A + B \right) \right\}\sin\frac{C}{2}\sin\left( \frac{A - B}{2} \right) \right] \left( \because A + B + C = \pi \right)\]
\[ = k\left[ \sin\left( B + C \right)\sin\frac{A}{2}\sin\left( \frac{B - C}{2} \right) + \sin\left( A + C \right)\sin\frac{B}{2}\sin\left( \frac{C - A}{2} \right) + \sin\left( A + B \right)\sin\frac{C}{2}\sin\left( \frac{A - B}{2} \right) \right]\]
\[ = k\left[ 2\sin\left( \frac{B + C}{2} \right)\cos\left( \frac{B - C}{2} \right)\sin\frac{A}{2}\sin\left( \frac{B - C}{2} \right) + 2\sin\left( \frac{A + C}{2} \right)\cos\left( \frac{C - A}{2} \right)\sin\frac{B}{2}\sin\left( \frac{C - A}{2} \right) + 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)\sin\frac{C}{2}\sin\left( \frac{A - B}{2} \right) \right]\]
\[ = 2k\left[ \sin\left( \frac{B + C}{2} \right)\sin\frac{A}{2}\sin\frac{A}{2}\sin\left( \frac{B - C}{2} \right) + \sin\left( \frac{A + C}{2} \right)\sin\frac{B}{2}\sin\frac{B}{2}\sin\left( \frac{C - A}{2} \right) + \sin\left( \frac{A + B}{2} \right)\sin\frac{C}{2}\sin\frac{C}{2}\sin\left( \frac{A - B}{2} \right) \right]\]
\[ = 2k\left[ \sin\left( \frac{B + C}{2} \right)\sin\left( \frac{B - C}{2} \right) \sin^2 \frac{A}{2} + \sin\left( \frac{A + C}{2} \right)\sin\left( \frac{C - A}{2} \right) \sin^2 \frac{B}{2} + \sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right) \sin^2 \frac{C}{2} \right]\]
\[ = 2k \sin^2 \frac{A}{2}\left( \sin^2 \frac{B}{2} - \sin^2 \frac{C}{2} \right) + 2k \sin^2 \frac{B}{2}\left( \sin^2 \frac{C}{2} - \sin^2 \frac{A}{2} \right) + 2k \sin^2 \frac{C}{2}\left( \sin^2 \frac{A}{2} - \sin^2 \frac{B}{2} \right)\]
\[ = 2k\left( \sin^2 \frac{A}{2} \sin^2 \frac{B}{2} - \sin^2 \frac{A}{2} \sin^2 \frac{C}{2} + \sin^2 \frac{B}{2} \sin^2 \frac{C}{2} - \sin^2 \frac{A}{2} \sin^2 \frac{B}{2} + \sin^2 \frac{A}{2} \sin^2 \frac{C}{2} - \sin^2 \frac{C}{2} \sin^2 \frac{B}{2} \right)\]
\[ = k\left( 0 \right)\]
\[ = 0\]
Hence proved.
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