Verify the inequality (a-b)/sin^2a + sina/cosa

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) = cot 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/(sin
x)^2

f'(c) = -1/(sin c)^2

cot
b - cot a = (b-a)/(cot c)^2

-1/(sin c)^2 = (cot b - cot
a)/(b-a)

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

a<c<b => sin a < sin c
> sin b

We'll raise to
square:

(sin a)^2 <(sin c)^2 < (sin
b)^2

-1/(sin a)^2 < -1/(sin c)^2 <-1/(sin
b)^2 (2)

But -1/(sin c)^2 = (cot b - cot a)/(b-a)
(1)

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

-1/(sin a)^2 <
(cot b - cot a)/(b-a) < -1/(sin
b)^2

We'll multiply by
(b-a):

(a-b)/(sin a)^2 <
cot b - cot a < (a-b)/(sin
b)^2

(a-b)/(sin a)^2 + cot a > cot b >
(a-b)/(sin b)^2 + cot a

(a-b)/(sin a)^2 +
cos a/sina < cot b < (a-b)/(sin b)^2 + cos a/sin
a

As we can notice, the given inequality is
not verified.