With usual notations in triangle ABC,if frac{ | vedclass

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(C) Given $\frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)}$.Applying cross-multiplication: $\sin A \sin (B-C) = \sin C \sin (A-B)$.Using the exp

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$AP$

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(C) Given $\frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)}$.Applying cross-multiplication: $\sin A \sin (B-C) = \sin C \sin (A-B)$.Using the expansion $\sin(x-y) = \sin x \cos y - \cos x \sin y$:$\sin A (\sin B \cos C - \cos B \sin C) = \sin C (\sin A \cos B - \cos A \sin B)$.$\sin A \sin B \cos C - \sin A \cos B \sin C = \sin C \sin A \cos B - \sin C \cos A \sin B$.Rearranging terms: $\sin A \sin B \cos C + \sin C \cos A \sin B = 2 \sin A \cos B \sin C$.$\sin B (\sin A \cos C + \cos A \sin C) = 2 \sin A \cos B \sin C$.Since $\sin A \cos C + \cos A \sin C = \sin(A+C) = \sin(\pi - B) = \sin B$,we have:$\sin B \cdot \sin B = 2 \sin A \cos B \sin C$.$\sin^2 B = 2 \sin A \sin C \cos B$.Dividing by $\sin A \sin C \sin B$: $\frac{\sin B}{\sin A \sin C} = \frac{2 \cos B}{\sin B} = 2 \cot B$.Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.Substituting these: $\frac{(b/2R)^2}{(a/2R)(c/2R)} = 2 \cos B \Rightarrow \frac{b^2}{ac} = 2 \cos B$.Using the cosine rule $\cos B = \frac{a^2+c^2-b^2}{2ac}$:$\frac{b^2}{ac} = 2 \left( \frac{a^2+c^2-b^2}{2ac} \right) = \frac{a^2+c^2-b^2}{ac}$.$b^2 = a^2+c^2-b^2 \Rightarrow 2b^2 = a^2+c^2$.Thus,$a^2, b^2, c^2$ are in $AP$.

With usual notations in $\triangle ABC$,if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$,then $a^2, b^2, c^2$ are in

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