using lmvt prove that tan x>x in (0,pi/2)

using lmvt prove that tan x>x in (0,pi/2)


2 Answers

askIITians Faculty 74 Points
9 years ago
Hello student,
tan x > xis equivalent to
tan x - x > 0.
Let therefore
f(x) = tan x - x .
We want to show that f(x)>0 on (0,pi/2).
It is clear that f(0) = 0, and by the mean-value theorem you get for any b in (0,pi/2) and for some z in (0,b),
f(b) - f(0) = f'(z) (b - 0) ,
which is equivalent to
f(b) = f'(z) * b .
Since f'(z)=(1-cos^2(x))/cos^2(x) >0 for all z in (0,pi/2), f(b) must be greater than zero on (0,pi/2), since the RHS above is a product of two positive numbers.
Thanks and Regards
Shaik Aasif
askIITians faculty
Dinesh Kumar
26 Points
one year ago
Let f(x)= tanx-x. = F`(x)=sec^2x-1 >0 
  Foe every x belongs to (0,π/2) 
  Since f(x) is increasing foe every 
  X belongs to (0,π/2)
  Now , x>0 = f(x)>f(0)
  = F(0)>0. ( Since , f(0) = tan0-0 =0-0=0)
 : Tanx-x>0
     Therefore tanx>x foe every x belongs to (0,π/2)

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