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In any DeltaABC, prove that "a sin (B - ...

In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A - B)=0`.

A

0

B

`sinA+sinB+sinC`

C

`sin^(2) A + sin^(2) B + sin^(2) C`

D

abc

Text Solution

Verified by Experts

The correct Answer is:
A
By sine rule ,we have
`(sinA)/(a)=(sinB)/(b)=(sinC)/(c)=k` (say)
Therefore , sin A=ak , sin B=bk , sin C = ck
Now , a `sin(B-C)+bsin(C-A)+csin(A-B)`
`=ksinAsin(B-C)+ksinBsin(C-A)+ksinCsinA-B)`
`=k[sin(B+C)sin(B-C)+sin(C+A)sin(C-A)+sin(A+B)sin(A-B)]`
`[becausesinA=sin{180^(@)-(B+C)}=sin(B+C)]`
`=k[sin^(2)B-sin^(2)C+sin^(2)C-sin^(2)A+sin^(2)A-sin^(2)B]`
`[becausesin(x+y)sin(x-y)=sin^(2)x-sin^(2)y]`
=k (0) =0
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