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

Answer

$\frac{\text{a}^2-\text{c}^2}{\text{b}^2}=\frac{\sin(\text{A}-\text{C})}{\sin(\text{A}+\text{C})}$ let $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$ $\text{LHS}=\frac{\text{a}^2-\text{c}^2}{\text{b}^2}$ $=\frac{\text{k}^2\sin^2\text{A}-\text{k}^2\sin^2\text{C}}{\text{k}^2\sin^2\text{B}}$ $=\frac{\text{k}^2(\sin^2\text{A}-\sin^2\text{C})}{\text{K}^2\sin^2\text{B}}$ $=\frac{(\sin^2\text{A}-\sin^2\text{C})}{\sin^2(\pi-(\text{A + C})}$ $=\frac{\sin(\text{A + C})\sin(\text{A} - \text{C})}{\sin^2(\text{A + C})}$ $=\frac{\sin(\text{A} - \text{C})}{\sin\text{(A + C)}}=\text{RHS}$

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