Explanation

(c)
Put $x = \tan{\theta\ \Rightarrow \ \ \ \theta = \tan^{- 1}x}$
As $x \rightarrow 0 \Rightarrow \theta \rightarrow 0$
$\therefore\ \ \lim_{\theta \rightarrow 0}{\frac{1}{\tan\theta}\sin^{- 1}\left( \frac{2\tan\theta}{1 + \tan^{2}\theta} \right)}$
$= \lim_{\theta \rightarrow 0}\ \frac{1}{\tan\theta}\sin^{- 1}{(\sin{2\theta})}$
$= \lim_{\theta \rightarrow 0}\frac{2\theta}{\tan\theta} = 2$