On simplifying $\cot \left(2 \tan ^{-1} x\right)$ where $|x|<1$, we get:

$\frac{1-x^2}{2 x}$ $\frac{2 x}{1-x^2}$ $\frac{1+x^2}{2 x}$ $\frac{2 x}{1+x^2}$

$\frac{1-x^2}{2 x}$

The correct answer is Option (1) → $\frac{1-x^2}{2 x}$