If in triangle ABC,with usual notations,a^2, | vedclass

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(B) Given that $a^2, b^2, c^2$ are in $A$.$P$.,we have $2b^2 = a^2 + c^2$. Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we get $\frac{\sin 3B}{\

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$\left(\frac{a^2-c^2}{2ac}\right)^2$

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(B) Given that $a^2, b^2, c^2$ are in $A$.$P$.,we have $2b^2 = a^2 + c^2$. Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we get $\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B$. Substituting $\sin^2 B = 1 - \cos^2 B$,we have $3 - 4(1 - \cos^2 B) = 4\cos^2 B - 1$. Using the cosine rule $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,we substitute $a^2 + c^2 = 2b^2$: $\cos B = \frac{2b^2 - b^2}{2ac} = \frac{b^2}{2ac}$. Thus,$4\cos^2 B - 1 = 4\left(\frac{b^2}{2ac}\right)^2 - 1 = \frac{4b^4}{4a^2c^2} - 1 = \frac{b^4 - a^2c^2}{a^2c^2}$. Since $b^2 = \frac{a^2 + c^2}{2}$,then $b^4 = \frac{(a^2 + c^2)^2}{4}$. Substituting this: $\frac{\frac{(a^2 + c^2)^2}{4} - a^2c^2}{a^2c^2} = \frac{(a^2 + c^2)^2 - 4a^2c^2}{4a^2c^2} = \frac{(a^2 - c^2)^2}{4a^2c^2} = \left(\frac{a^2 - c^2}{2ac}\right)^2$.

If in $\triangle ABC$,with usual notations,$a^2, b^2, c^2$ are in $A$.$P$.,then $\frac{\sin 3B}{\sin B} =$

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