In a Delta ABC, if frac{sin A}{sin C} = frac{ | vedclass

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(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

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$A.P.$

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(A) Given $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)}$.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,we get $\frac{a}{c} = \frac{\sin A \cos B - \cos A \sin B}{\sin B \cos C - \cos B \sin C}$.Using the projection formula $a = b \cos C + c \cos B$,we have $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Cross-multiplying: $a(\sin B \cos C - \cos B \sin C) = c(\sin A \cos B - \cos A \sin B)$.Using $\sin A = \frac{a}{2R}$,etc.,this simplifies to $a(b \cos C - c \cos B) = c(a \cos B - b \cos A)$.$ab \cos C - ac \cos B = ac \cos B - bc \cos A$.$ab \cos C + bc \cos A = 2ac \cos B$.Substituting $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and $\cos A = \frac{b^2+c^2-a^2}{2bc}$ and $\cos B = \frac{a^2+c^2-b^2}{2ac}$:$ab(\frac{a^2+b^2-c^2}{2ab}) + bc(\frac{b^2+c^2-a^2}{2bc}) = 2ac(\frac{a^2+c^2-b^2}{2ac})$.$\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} = a^2+c^2-b^2$.$\frac{2b^2}{2} = a^2+c^2-b^2$.$b^2 = a^2+c^2-b^2 \implies 2b^2 = a^2+c^2$.Thus,$a^2, b^2, c^2$ are in $A.P.$

In a $\Delta ABC,$ 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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