Using Lagrange’s mean value theorem, prove that (b – a ) sec 2 a

Question

Using Lagrange’s mean value theorem, prove that (b – a ) sec 2 a < tan b – tan a < (b – a) sec 2 b, where 0 < a < b < π/2.

Philoid · Accepted Answer

Let f(x) = tan x on [a, b]

We know that, tan x function is continuous and differentiable on

differentiable on (a, b).

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = tan x

Differentiating with respect to x:

f’(x) = sec² x

Since c lies between a and b

⇒ a < c < b

⇒ sec² a < sec² c < sec² b

⇒ sec² a (b – a)2 b (b – a)

Hence Proved

Using Lagrange’s mean value theorem, prove that(b – a ) sec2 a < tan b – tan a < (b – a) sec2 b, where 0 < a < b < π/2.

Using Lagrange’s mean value theorem, prove that
(b – a ) sec² a < tan b – tan a < (b – a) sec² b, where 0 < a < b < π/2.