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

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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.
answered Feb 19, 2013 by thanvigandhi_1
edited Mar 13, 2013 by rvidyagovindarajan_1
 

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