MCQ
If $A = {\tan ^{ - 1}}x$, then $\sin 2A = $
  • A
    $\frac{{2x}}{{\sqrt {1 - {x^2}} }}$
  • B
    $\frac{{2x}}{{1 - {x^2}}}$
  • $\frac{{2x}}{{1 + {x^2}}}$
  • D
    None of these

Answer

Correct option: C.
$\frac{{2x}}{{1 + {x^2}}}$
c
(c) Given that $A = {\tan ^{ - 1}}x$

Now $x = \tan A \Rightarrow \sin 2A = \frac{{2\tan A}}{{1 + {{\tan }^2}A}} = \frac{{2x}}{{1 + {x^2}}}$.

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