Prove that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))

**verasi** · February 15th 2013, 09:52 PM

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

**ibdutt** · February 15th 2013, 10:24 PM

Thread: Prove that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))

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Prove that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))

Re: Prove that arctan(a)-arctan(c)=arctan((a-b)/(1+ab))+arctan((b-c)/(1+bc))

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