We have,
$$
\begin{aligned}
& \int \frac{2 d x}{\sqrt{\cot ^2 x-\tan ^2 x}} d x=\int \frac{2 d x}{\sqrt{\frac{\cos ^2 x}{\sin ^2 x}-\frac{\sin ^2 x}{\cos ^2 x}}} \\
& =\int \frac{2 \sin x \cos x}{\sqrt{\cos ^4 x-\sin ^4 x}} d x \quad[\because \sin 2 x=2 \sin x \cos x] \\
& =\int \frac{\sin 2 x d x}{\sqrt{\left(\cos ^2 x-\sin ^2 x\right)\left(\cos ^2 x+\sin ^2 x\right)}} \\
& \quad\left[\because \sin ^2 x+\cos ^2 x=1 \operatorname{and~}^2 x-\sin ^2 x=\cos 2 x\right] \\
& =\int \frac{\sin 2 x d x}{\sqrt{\cos 2 x}} \\
& \text { Put } \cos 2 x=t \quad \Rightarrow \quad-2 \sin 2 x d x=d t \\
& \Rightarrow-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \frac{t^{\frac{-1}{2}}+1}{-\frac{1}{2}+1}+c \Rightarrow-\frac{1}{2} \frac{t^{1 / 2}}{\frac{1}{2}}+c=-\sqrt{t}+c \\
& \quad=-\sqrt{\cos 2 x}+c
\end{aligned}
$$
$[t=\cos 2 x]$
$$
\begin{gathered}
=-\sqrt{\cos 2 x}+c \\
\text { Compare with }-\sqrt{f(x)}+c
\end{gathered}
$$

So, $f(x)=\cos 2 x$