Math, asked by routl856, 6 months ago

Prove that
cos A sin (B-C) + sin B sin (C - A) + cos C sin (A-B) = 0. ​

Answers

Answered by hanockgamer611
0

Step-by-step explanation:

ANSWER

Use sin formula

asinA=bsinB=csinC=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

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