# Prove that (b-a)/cos^2 a < tan b - tan a < (b-a)/cos^2 b if 0=<a<b<pi/2.

*print*Print*list*Cite

### 1 Answer

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**