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

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