In a `DeltaABC`
`(bsin(C-A))/(c^(2)-a^(2))+(csin(A-B))/(a^(2)-b^(2))=`

Options

A. `1/(2R)`
B. `1/R`
C. `2/R`
D. none of these

Explanation

We have,
`a/(2sinA)=b/(2sinB)=c/(2sinC)=R`
`therefore(bsin(C-A))/(c^(2)-a^(2))+c(sin(A-B))/(a^(2)-b^(2))`
`=(2RsinBsin(C-A))/(4R^(2)(sin^(2)C-sin^(2)A))+(2RsinCsin(A-B))/(4R^(2)(sin^(2)A-sin^(2)B))`
`=1/(2R){(sin(C+A)sin(C-A))/(sin(C+A)sin(C-A))+(sin(A+B)sin(A-B))/(sin(A+B)sin(A-B))}`
`=1/(2R)xx2=1/R`