Question

Prove a sin(b-c)+b sin(c-a) + c sin (a-b)=0

Answer 1

Answer:

Prove that asin(B−C)+bsin(C−A)+csin(A−B)=0

Use sin formula

$\bold{\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k}$

sinA=ak,sinB=bk,sinc=ck

Also, sin(A−B)=sinA..cosB−cosA−sinB

=akcosB−cosA.bk

k(acosB−bcosA)

Similarity, sin(B−C)=k(bcosC−ccosB)

sin(C−A)=k(ccosA−acosc)

LHS=asin(B−C)+bsin(C−A)+csin(A−B)

=ak(bcosc−cosB+bk(ccosA−acosC+ck(acosB−bcosA)

=k(bccosA−bccosA)+k(accosB−accosB)+(abcos−abcosc)

=0+0+0

=RHS

Answer 2

Answer:

answer to your question is in the following pic

Step-by-step explanation:

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