Explanation

(c)
Given, $\cos^{- 1}{x = \alpha}$
$\Rightarrow \ \ x = \cos{\alpha,\ \ \ 0 < x < 1}$ ...(i)
Also, $\sin^{- 1}{\left( 2x\sqrt{1 - x^{2}} \right) + \sec^{- 1}{\left( \frac{1}{2x^{2} - 1} \right) = \frac{2\pi}{3}}}$
$\therefore\ \sin^{- 1}{\left( 2\cos{\alpha\sqrt{1 - \cos^{2}\alpha}} \right) + \sec^{- 1}\left( \frac{1}{2\cos^{2}{\alpha - 1}} \right)} = \frac{2\pi}{3}$
$\Rightarrow \ \ \sin^{- 1}{\left( \sin{2\alpha} \right) + \sec^{- 1}{\left( \sec{2\alpha} \right) = \frac{2\pi}{3}}}$
$\Rightarrow \ \ 2\alpha + 2\alpha = \frac{2\pi}{3}$
l$\Rightarrow \ \ \ \alpha = \frac{\pi}{6}$
Now, $x = \cos{\frac{\pi}{6} = \frac{\sqrt{3}}{2}}$
$\Rightarrow \ \ 2x = \sqrt{3}$
$\therefore\ \ \ \tan^{- 1}{(2x) = \tan^{- 1}{\left( \sqrt{3} \right) = \frac{\pi}{3}}}$