Question #d4c2f

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Answer 1 · 2016-08-28T04:36:12

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