Using Lagranges mean value theorem, prove that `(b-a)sec^2a lt (tanb-tana) lt (b-a)sec^2b` , where `0ltaltbltpi/2`

Explanation

Lagranges 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 [a,b]` such that `0 lt a lt b lt pi/2`


Then, from Lagranges Mean value theorem,

`sec^2 c = (tanb-tana)/(b-a)->(1)`

Here, `a lt c lt b`

`=>sec^2 a lt sec^2c lt sec^2b`

...