Simplify: $\;$ $\cos \left(2 \tan^{-1} x\right)$
Given expression: $\;$ $\cos \left(2 \tan^{-1} x\right)$
$= 2 \cos^2 \left(\tan^{-1} x\right) - 1$ $\;\;\; \left\{\because \; \cos 2 \theta = 2 \cos^2 \theta - 1\right\}$
$= 2 \left[\cos \left(\tan^{-1} x\right)\right]^2 - 1$
$= 2 \times \left(\dfrac{1}{\sqrt{1 + x^2}}\right)^2 - 1$ $\;\;\; \left\{\because \; \cos \left(\tan^{-1} x \right) = \dfrac{1}{\sqrt{1 + x^2}}, \; x > 0\right\}$
$= \dfrac{2}{1 + x^2} - 1$
$= \dfrac{1 - x^2}{1 + x^2}$