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

  • A
    $\frac{a^2-c^2}{2ac}$
  • B
    $\left(\frac{a^2-c^2}{2ac}\right)^2$
  • C
    $\frac{a^2-c^2}{ac}$
  • D
    $\left(\frac{a^2-c^2}{ac}\right)^2$

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