a^2 (B-C)=(b^2-c^2) a = b = c =k LHS=a^2 (B-C) =a^2( . - . ) =a^2kb. a^2+b^2-c^2 2ab -a^2ck. a^2+c^2-b^2 2ac [Using cos rule and sine rule] =a^2k. a^2+b^2-c^2 2a -a^2k. a^2+c^2-b^2 2a =a^2k. ( a^2+b^2-c^2-a^2-c^2+b^2 2a ) =a^2k. ( 2b^2-2c^2 2a ) =ak.(b^2-c^2) = (b^2-c^2)=RHS Hence Proved

a^2 (B-C)=(b^2-c^2)

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

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Answer

$\text{a}^2\sin(\text{B}-\text{C})=(\text{b}^2-\text{c}^2)\sin\text{A}$
$\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\text{LHS}=\text{a}^2\sin(\text{B}-\text{C})$
$=\text{a}^2(\sin\text{B}.\cos\text{C}-\sin\text{C}.\cos\text{B})$
$=\text{a}^2\text{kb.}\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}-\text{a}^2\text{ck.}\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$ [Using cos rule and sine rule]
$=\text{a}^2\text{k}.\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{a}}-\text{a}^2\text{k}.\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}}$
$=\text{a}^2\text{k}.\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{a}^2-\text{c}^2+\text{b}^2}{2\text{a}}\Big)$
$=\text{a}^2\text{k}.\Big(\frac{2\text{b}^2-2\text{c}^2}{2\text{a}}\Big)$
$=\text{ak.(b}^2-\text{c}^2)$
$=\sin\text{A(b}^2-\text{c}^2)=\text{RHS}$
Hence Proved

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