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

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

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