**Question:**

In triangle ABC, prove the following:

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

**Solution:**

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $\frac{a^{2}-c^{2}}{b^{2}}=\frac{\sin (A-C)}{\sin (A+C)}$.

$\mathrm{LHS}=\frac{[k \sin (\mathrm{A})]^{2}-[\mathrm{k} \sin (\mathrm{C})]^{2}}{(\mathrm{k} \sin (\mathrm{B}))^{2}}$

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

$=\frac{\sin (A+C) \sin (A-C)}{\sin ^{2} B}$ $\left[\because \sin ^{2} A-\sin ^{2} C=\sin (A+C) \sin (A-C)\right]$

$=\frac{\sin (A+C) \sin (A-C)}{\operatorname{Sin}^{2}(\pi-(\mathrm{A}+\mathrm{C}))}$ $[\because(A+B+C)=\pi]$

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

$=\frac{\sin (A-C)}{\sin (A+C)}=\mathrm{RHS}$

Hence proved.