Range of f (x)=secx + tanx- 1/tanx-secx+1, x belongs to (0, pi/2) is - Maths - Inverse Trigonometric Functions

Question

Range of f (x)=secx + tanx- 1/tanx-secx+1, x belongs to (0,
pi/2) is

Aarushi Mishra · Accepted Answer

fx=sec x+tan x-1tan x-sec x+1fx=sec x+tan x-1tan x-sec x-1sec2x-tan2x=1sec x-tan xsec x+tan x=1sec x-tan x=1sec x+tan xfx=sec x+tan x-11sec x+tan x+1Let sec x+tan x=ufx=u-11u+1fx=uu-1u+1u=sec x+tanxu'=secx tanx +sec2x>0 in 0,π2u is increasing function at x=0, u=1 and x→π2, u→∞Note for u>1, fx will always be greater than zero and as u→∞, fx will go to ∞Hence range of fx is 0,∞