In any ΔABC, prove that

Solution:

Need to prove: $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\left(a^{2}-b^{2}\right)}{c^{2}}$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius.

Therefore,

$a=2 R \sin A \cdots(a)$

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

From Right hand side,

$=\frac{a^{2}-b^{2}}{c^{2}}$

$=\frac{4 R^{2} \sin ^{2} A-4 R^{2} \sin ^{2} B}{4 R^{2} \sin ^{2} C}$

$=\frac{4 R 2\left(\sin ^{2} A-\sin ^{2} B\right)}{4 R^{2} \sin ^{2} C}$

$=\frac{\sin (A+B) \sin (A-B)}{\sin ^{2} C}$

$=\frac{\sin (A+B) \sin (A-B)}{\sin ^{2}(\pi-(A+B))}$

$=\frac{\sin (A+B) \sin (A-B)}{\sin ^{2}(A+B)}$

$=\frac{\sin (A-B)}{\sin (A+B)}$

$=$ Left hand side. [Proved]