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) sec x(sec x – tan x) sec x(tan x + sec x) none of these

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.