Question

Show that Σsin (b-c) /sin b sin c = 0​

Answer 1

∑a(sinB−sinC)=0

L.H.S.−

∑a(sinB−sinC)=a(sinB−sinC)+b(sinC−sinA)+c(sinA−sinB)

From sin rule, i.e.,

$\sf \small[ \frac{sin \: a}{a} = \frac{sin \: b}{b} = \frac{sin \: c}{c} = k ]$

∑a(sinB−sinC)=a(bk−ck)+b(ck−ak)+c(ak−bk)

⇒∑a(sinB−sinC)=abk−ack+bck−abk+cak−cbk

⇒∑a(sinB−sinC)=0=R.H.S.

Hence proved!!

Answer 2

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∑a(sinB−sinC)=0

L.H.S.−

$∑a(sinB−sinC)=a(sinB−sinC)+b(sinC−sinA)+c(sinA−sinB)$

From the rule that ie..

$\sf \small[ \frac{sin \: a}{a} = \frac{sin \: b}{b} = \frac{sin \: c}{c} = k ]$

∑a(sinB−sinC)=a(bk−ck)+b(ck−ak)+c(ak−bk)

⇒∑a(sinB−sinC)=abk−ack+bck−abk+cak−cbk

⇒∑a(sinB−sinC)=0=R.H.S.