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

We'll use Lagrange's theorem to prove the trigonometric
inequality.

We'll choose a function, whose domain of
definition is the closed interval [a,b].

The function is
f(x) = tan x

Based on Lagrange's theorem, there is a point
"c", that belongs to (a,b), so that:

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

We'll substitute the function f(x) in the relation
above:

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

We'll determine
f'(x):

f'(x) = 1/(cos
x)^2

f'(c) = 1/(cos c)^2

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

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

Since a and b are located in the interval [0 ;
pi/2], the cosine function over this interval is decreasing and it has positive
values.

a<c<b => cos a > cos c
> cos b

We'll raise to
square:

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

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

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

From (1) and (2), we'll
get:

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