Answer
$\dfrac{2 \sec^2 x \tan x}{(1+\sec^4 x)^{3/2}}$
Work Step by Step
Given: $y=\tan x$
Consider $f(x)=y=\tan x$
In order to find the curvature we will have to use formula 11, such that
$\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
$y'=\sec^2 x$ and $y''=2 \sec^2 x \tan x$
$|y''|=2 \sec^2 x \tan x$
$\kappa(x)=\dfrac{|2 \sec^2 x \tan x|}{[1+(\sec^2 x)^2]^{3/2}}$
Hence, $\kappa(x)=\dfrac{2 \sec^2 x \tan x}{(1+\sec^4 x)^{3/2}}$
