Prove that (b-a)/cos^2a

To prove this inequality, we'll consider a function f(x) =
tan x, whose domain of definition is [0,pi/2). The values a and b are included in this
interval.

Since the function tan x is continuous and
differentiable over the interval [a,b], we could apply Lagrange's
theorem:

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

tan b - tan a = (b-a)/(cos
c)^2

The c value is included in the interval [a,b], such
as:

a < c <
b

Since the cosine function is decreasing over [0,pi/2),
we'll get:

cos a > cos c > cos
b

(cos a)^2 > (cos c)^2 > (cos
b)^2

1/(cos a)^2 < 1/(cos c)^2 < 1/(cos
b)^2

But 1/(cos c)^2 = (tan b - tan
a)/(b-a)

The inequality will
become:

1/(cos a)^2 < (tan b - tan a)/(b-a) <
1/(cos b)^2

Since the value of the difference b - a is
positive, if we'll multiply the inequality by (b-a), it won't
change:

(b-a)/(cos a)^2 < (b-a)*(tan b - tan
a)/(b-a) < (b-a)/(cos b)^2

We'll simplify and we'll
get:

(b-a)/(cos a)^2 < (tan b - tan a)/(b-a)
< (b-a)/(cos b)^2

We notice that
applying Lagrange's theorem, the inequality (b-a)/(cos a)^2 < (tan b - tan
a)/(b-a) < (b-a)/(cos b)^2 is
verified.