Show that tanx>x for 0<x<pi/2

By the Mean Value Theorem, if f is continuous on [a, b] and differentiable on (a, b),

(f(b) - f(a))/(b - a) = f'(c) for some c with a < c < b. (*)

Let f(t) = tan t, a = 0, and b = x. Then the derivative of f(t), f'(t), is (sec t)^2, so (*) becomes

(tan x)/x = (sec c)^2, for some c with 0 < c < x < π/2.

However, sec t exceeds 1 on (0, π/2), so (sec c)^2 > 1. This means that (tan x)/x > 1, so tan x > x. This completes the proof.

Substitute x as 3 in radians or 30 in degrees and NEITHER is greater than x.

These two counter-examples show your statement is FALSE

By chance did you mean....
SHOW TAN(X)>0 for 0
In the FIRST quadrant of the UNIT CIRCLE all x and y coordinates are POSITIVE
Tangent ration is the y-value divided by the x-value
Positive divided by a positive is positive
Positives are greater 0.........>0
thus this new statement is TRUE

Let f(x) = tan(x)-x
f'(x) = sec^2(x)-1
For the interval (0,pi/2)
sec^2(x) is >1
f'(x) is positive
tan(x)-x>0
tan(x)>x

See:
http://studysphere.ru/image/HigherMathem…
MK=sin x, AN=tg x


Rsin(x)<2πR*x/(2π) sin(x)