If $\tan x = \frac{2b}{a - c}$ $(a \ne c)$,$y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$,then:

  • A
    $y = z$
  • B
    $y + z = a + c$
  • C
    $y - z = a + c$
  • D
    $y - z = (a - c)^2 + 4b^2$

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