show by using mean value theorem the B-A/1+B^2 tan^-1B - tan^-1A B-A/1+A^2,where B A 0 - Maths - Application of Derivatives

Question

show by using mean value theorem the B-A/1+B^2 < tan^-1B - tan^-1A < B-A/1+A^2,where B>A>0

Anuradha Sharma · Accepted Answer

Dear Student,
Let f(x) = arctan x for x in [a, b], a subset of (0, 1).

Then, the Mean Value Theorem yields
(arctan b - arctan a)/(b - a) = 1/(1 + c^2) for some c in (a, b).

Since g(t) = 1/(1 + t^2) is decreasing for t > 0, we have
1/(1 + b^2) < 1/(1 + c^2) < 1/(1 + a^2).

Hence,
1/(1 + b^2) < (arctan b - arctan a)/(b - a) < 1/(1 + a^2).
==> (b - a)/(1 + b^2) < arctan b - arctan a < (b - a)/(1 + a^2).

Regards