Question

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

• A. True
• B. False

Answer 1

The correct option is A True

In $\Delta ABC$

From $\sin$ rule

$a=2R\sin A,b=2R\sin B,c=2R\sin C$

Consider

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

Multiplying and dividing by $\sin (A+B)$

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

$⟹\frac{({\sin }^{2}A-{\sin }^{2}B)}{{\sin }^2(\pi -C)}$

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

$⟹\frac{a^2-b^2}{c^2}$