In Delta ABC,if sin A : sin C = sin (A - B) : | vedclass

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(B) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.Since $A + B + C = \pi$,we have $A = \pi - (B + C)$,so $\sin A = \sin (B + C)$.S

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$a^2, b^2, c^2$ are in $A.P.$

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(B) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.Since $A + B + C = \pi$,we have $A = \pi - (B + C)$,so $\sin A = \sin (B + C)$.Similarly,$C = \pi - (A + B)$,so $\sin C = \sin (A + B)$.Substituting these,we get $\frac{\sin (B + C)}{\sin (A + B)} = \frac{\sin (A - B)}{\sin (B - C)}$.Cross-multiplying gives $\sin (B + C) \sin (B - C) = \sin (A + B) \sin (A - B)$.Using the identity $\sin (x+y) \sin (x-y) = \sin^2 x - \sin^2 y$,we get $\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$.Rearranging,$2 \sin^2 B = \sin^2 A + \sin^2 C$.By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.Substituting these,$2(\frac{b}{2R})^2 = (\frac{a}{2R})^2 + (\frac{c}{2R})^2$.This simplifies to $2b^2 = a^2 + c^2$.Thus,$a^2, b^2, c^2$ are in $A.P.$

In $\Delta ABC$,if $\sin A : \sin C = \sin (A - B) : \sin (B - C)$,then

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