If $0 < B < A < \frac{\pi}{4}$,$\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$,then $\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} =$

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

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