Answer
$d$
Work Step by Step
$\theta = \tan^{-1} {x} $
$\tan{\theta} = x$
$\sec{\theta} = \sqrt{1+\tan^2{\theta}} = \sqrt{1+x^2}$
$\cos{\theta} =\dfrac{1}{\sec{\theta}} = \dfrac{1}{\sqrt{1+x^2}}$
$\sin{\theta} = \sqrt{1-\cos^2{\theta}} = \dfrac{x}{\sqrt{1+x^2}}$
$\sin{2\theta} = 2 \sin{\theta} \cos{\theta} = \dfrac{2x}{1+x^2}$