In $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{(\sin^2 A + \sin^2 B)}{(\sin^2 A - \sin^2 B)} \sin(A - B) = $

  • A
    $1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{\sqrt{3}}{2}$
  • D
    $0$

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