In `DeltaABC`, prove that: `asin(B-C)+bsin(C-A)+csinA-B)=0`

1.	In `DeltaABC`, prove that: `asin(B-C)+bsin(C-A)+csinA-B)=0`
Answer» 1st term =`asin(B-C)` `=ksinA.sin(B-C)` `=ksin{180^(@)-(B+C)}sin(B-C)` `ksin(B+C).sin(B-C)` `=ksin^(2)B-sin^(2)C` Similarly, 2nd term `=k(sin^(2)C-sin^(2)A)` 3rd term `=k(sin^(2)A-sin^(2)B)` Now L.H.S. `=asin(B-C)+bsin(C-a)+csin(A-B)` `=k(sin^(2)B-sin^(2)C)+k(sin^(2)C-sin^(2)A)+k(sin^(2)A-sin^(2)B)` `ksin^(2)B-ksin^(2)C-ksin^(2)A+ksin^(2)A-ksin^(2)B` =0 = RHS Hence Proved.

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