# Prove that :$\cot^{-1}\bigg[\frac{ab+1}{a-b}\bigg]+\cot^{-1}\bigg[\frac{bc+1}{b-c}\bigg]+\cot^{-1}\bigg[\frac{ca+1}{c-a}\bigg]=0$

Toolbox:
• $$cot^{-1}x=tan^{-1}\frac{1}{x}$$
• $$tan^{-1}tanx=x$$
• $$tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$$
L.H.S.
Using the above formula of $$cot^{-1}x$$ we can write L.H.S. as
$$tan^{-1}\frac{a-b}{1+ab}+tan^{-1}\frac{b-c}{1+bc}+tan^{-1}\frac{c-a}{1+ab}$$
$$Take \: a=tanA$$
$$b = tanB$$
$$c = tanC$$ Then
$$\frac{a-b}{1+ab}=\frac{tanA-tanB}{1+tanAtanB}= tan(A-B)$$
$$\frac{b-c}{1+bc}=\frac{tanB-tanC}{1+tanBtanC}= tan(B-C)$$
$$\frac{c-a}{1+ca}=\frac{tanC-tanA}{1+tanCtanA}= tan(C-A)$$
Substituting the values in L.H.S. we get
$$tan^{-1}\: tan(A-B)+tan^{-1}\: tan(B-C)+tan^{-1}\: tan(C-A)$$
$$= A-B+B-C+C-A=0$$
= R.H.S.
edited Mar 13, 2013