The correct option is C f(x) is an onto function
f(x)=[ln (sec x+tan x)]3f′(x)=3[ln (sec x+tan x)]2(sec x tan x+sec2x)(sec x+tan x)
f′(x)=3secx[ln(sec x+tan x)]2>0, ∀x ϵ (−π2, π2)
f(x) is an increasing function.
∴ f(x) is an one – one function.
(sec x+tan x)=tan(π4+x2), as x ϵ(−π2, π2), then 0<tan (π2+x2)<∞0<sec x+tan x<∞⇒ −∞<ln(sec x+tan x)<∞−∞<[ln (sec x+tan x)]3<∞⇒ −∞<f(x)<∞
Range of f(x) is R and thus f(x) is an onto function.
f(−x)=[ln(sec x−tan x)]3=[ln(1sec x+tan x)]3f(−x)=−[ln(sec x+tan x)]3
f(x) + f (-x) = 0
⇒ f(x) is an odd function.