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