$\displaystyle \arctan{a}-\arctan(b)=\arctan{frac{a-b}{1+ab}}+\arctan{frac{b-c}{1+bc}}$

I think the question has something to do with the tan(a+b) identity, but I'm not sure how to apply that identity here. Thanks for your help in advance :)

- Feb 15th 2013, 09:52 PMverasiProve that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))
$\displaystyle \arctan{a}-\arctan(b)=\arctan{frac{a-b}{1+ab}}+\arctan{frac{b-c}{1+bc}}$

I think the question has something to do with the tan(a+b) identity, but I'm not sure how to apply that identity here. Thanks for your help in advance :) - Feb 15th 2013, 10:24 PMibduttRe: Prove that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))