# Prove that (b-a)/cos^2a < tanb-tana < (b-a)/cos^2 b if 0=<a<b<1/pi/2

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### 1 Answer

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****.**