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If $ \tan^{-1}a+\tan^{-1}b+\tan^{-1}c=\pi,\;prove\;that\;a+b+c=abc.$

1 Answer

Toolbox:
  • \( tan^{-1}a+tan^{-1}b=tan^{-1}\large\frac{a+b}{1-ab}\)
  • \( tan (\pi\: -\theta))=-tan \theta\)
Given,\(tan^{-1}a+tan^{-1}b+tan^{-1}c=\pi\)
\(tan^{-1}a+tan^{-1}b=\pi-tan^{-1}c\)
 
Using the above formula of \( tan^{-1}a+tan^{-1}b\) we get
\( \Rightarrow tan^{-1} \large\frac{a+b}{1-ab}=\pi-tan^{-1}c\)
 
take tan on both sides
\(tan(tan^{-1} \large\frac{a+b}{1-ab})=tan(\pi- tan^{-1}c)\)
taking \(\theta=tan^{-1}c\) in the above formula we get
\( \large\frac{a+b}{1-ab}=-tan\: tan^{-1}c\)
\( \large\frac{a+b}{1-ab}=-c\)
 
\(\Rightarrow\:a+b=-c+abc\)
\( \Rightarrow a+b+c=abc\)

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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