જો $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,હોય,તો $x = $

  • A
    $a$
  • B
    $b$
  • C
    $\frac{{a + b}}{{1 - ab}}$
  • D
    $\frac{{a - b}}{{1 + ab}}$

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