# Verify the inequality (a-b)/sin^2a + sina/cosa<cot b<sina/cos a+(a-b)/sin^2b?

### 1 Answer | Add Yours

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** < c**ot 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.**