In any ΔABC, prove that sin(A - B)/sin(A + B) = (a^2 - b^2)/c^2

Question

In any ΔABC, prove that

\(\frac{sin(A -B)}{sin(A+B)}\) = \(\frac{(a^2-b^2)}{c^2}\)

Answer 1

To prove

\(\frac{sin(A -B)}{sin(A+B)}\) = \(\frac{(a^2-b^2)}{c^2}\)

We know that, a/sinA = b/sinB = csinC = 2R where R is the circumradius

Therefore, 

a = 2R sinA ---- (a) 

Similarly, b = 2R sinB and c = 2R sinC 

From Right hand side,

= \(\cfrac{a^2-b^2}{c^2}\)

= \(\frac{sin(A -B)}{sin(A+B)}\)

= Left hand side. [proved]