Question

Question #d4c2f

Answer 1

We know from properties of triangle that for `DeltaABC`

`a/sinA=b/sinB=c/sinC=2R...(1)`

where a,b,c are three sides of `DeltaABC` opposite to angles A,B,C respectively and R is the circum radius of `DeltaABC`

By relation (1)

`a/sinA=b/sinB=c/sinC=2R`

`=>a=2RsinA,b=2RsinB,c=2RsinC`

Now

`RHS=(a^2-b^2)/c^2=(4R^2(sin^2A-sin^2B))/(4R^2sin^2C)`

`=(sin(A+B)sin(A-B))/sin^2(pi-(A+B))`

`=(sin(A+B)sin(A-B))/sin^2(A+B)`
^
`=sin(A-B)/sin(A+B)=LHS`

Proved