If $f(x)=\frac{\tan x+\sec x-1}{\tan x-\sec x+1}$ then f’(x) is equal to

sec x(tan x + sec x)

$f(x)=\left(\frac{\tan x+\sec x-1}{\tan x-\sec x+1}\right) \times\left(\frac{\tan x+\sec x+1}{\tan x+\sec x+1}\right)$

$=\frac{\tan ^2 x+\sec ^2 x+2 \tan x \sec x-1}{(1+\tan x)^2-\sec ^2 x}=\tan x+\sec x$

Thus f’(x) = sec²x + sec x tan x = sec x (sec x + tan x)

Hence (3) is correct answer.