In an isosceles triangle $ABC, \angle C = \angle A$ if point of intersection of bisectors of internal angles $\angle A$ and $\angle C$ divide median of side $AC$ in $3 : 1$ (from vertex $B$ to side $AC$), then value of $cosec \ \frac{B}{2}$ is equal to

  • A

    $1$

  • B

    $2$

  • C

    $3$

  • D

    $4$

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