Anti-derivative of $\frac{\tan x - 1}{\tan x + 1}$ with respect to $x$ is:

$\log \left| \sec \left(\frac{\pi}{4} - x\right) \right| + c$

The correct answer is Option (3) → $\log \left| \sec \left(\frac{\pi}{4} - x\right) \right| + c$

Anti-derivative of $\frac{\tan x - 1}{\tan x + 1} = \int\frac{\tan x - 1}{\tan x + 1}dx$

$=\int\tan(x-\frac{\pi}{4})dx$

$=\int -\tan(\frac{\pi}{4}-x)dx$

$=\int -\tan(\frac{\pi}{4}-x)dx$

$=-\log\frac{\left| \sec \left(\frac{\pi}{4} - x\right) \right|}{-1}+c$

 $=\log \left| \sec \left(\frac{\pi}{4} - x\right) \right| + c$