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Prove that, (asin(B-C))/(b^(2)-c^(2)) ...

Prove that,
`(asin(B-C))/(b^(2)-c^(2)) = (bsin(C-A))/(c^(2)-a^(2)) = (csin(A-B))/(a^(2)-b^(2))`

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From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now, `(asin(B-C))/(b^2-c^2) = (ksinAsin(B-C))/(k^2(sin^2B-sin^2C))`
`= sin(pi-(B+C)sin(B-C))/(k(sin^2B-sin^2C))`
`= (sin(B+C)sin(B-C))/(k(sin^2B-sin^2C))`
We know, `sin(B+C)sin(B-C) = sin^2B-sin^2C`
So, it becomes,
...
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