In $\triangle ABC$,if $x=\tan \left(\frac{B-C}{2}\right) \tan \frac{A}{2}$,$y=\tan \left(\frac{C-A}{2}\right) \tan \frac{B}{2}$,and $z=\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$,then $(x+y+z)$ is equal to

  • A
    $xyz$
  • B
    $-xyz$
  • C
    $2xyz$
  • D
    $\frac{1}{2}xyz$

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