If $z=\sec (y-ax)+\tan (y+ax)$, then $\frac{\partial^2 z}{\partial x^2}-a^2 \frac{\partial^2 z}{\partial y^2}$ is equal to

  • A
    $0$
  • B
    $-z$
  • C
    $z$
  • D
    $2x$

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