Using mean value theorem, prove that `tanx > x` for all `x(0,pi/2)dot`

Explanation

Mean value theorem states that if a function `f(x)` is continuous and differentiable in interval `(a,b)`, then,

`(f(b)-f(a))/(b-a) = f'(c)`, where c lies in `(a,b)`.

Let `f(x) = tanx` where `x in (0,pi/2)` .


Then, from Lagranges Mean value theorem,

` f'(c) = (f(b)-f(a))/(b-a) `

If we apply Mean value theorem on `[0,x]`.

Then, `sec^2 c = (tanx-tan0)/(x-0)`

...