Prove $\sin(A-B)/\sin(A+B)=(a^2-b^2)/c^2$

Question

prove

For any triangle $\triangle ABC$, prove that

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

Answer 1

HINT:

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

and $\sin(A+B)=\sin(\pi-C)=?$

and Sine Law

Answer 2

$$\frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin A\cos B-\sin B\cos A}{\sin A\cos B+\sin B\cos A} = \frac{2ac\cos B-2bc\cos A}{2ac\cos B+2bc\cos A}=\frac{(a^2-b^2+c^2)-(-a^2+b^2+c^2)}{(a^2-b^2+c^2)+(-a^2+b^2+c^2)}=\frac{a^2-b^2}{c^2}.$$