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Prove the following : $ \tan^{-1} \bigg(\large\frac{1}{2} \bigg) + \tan^{-1} \bigg( \large\frac{1}{5} \bigg) + \tan^{-1} \bigg( \frac{1}{8} \bigg)=\large\frac{\pi}{4} $

1 Answer

Toolbox:
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg) , xy < 1\)
  • \(tan^{-1}1=\frac{\pi}{4}\)
By taking \(x=\frac{1}{2}\:and\:y=\frac{1}{5}\) in the formula of \( tan^{-1}x+tan^{-1}y\) we get
L.H.S.=
\( tan^{-1} \bigg[\large \frac{\large\frac{1}{2}+\large\frac{1}{5}}{1-\large\frac{1}{10}}\bigg] +tan^{-1}\large\frac{1}{8}\)
\(= tan^{-1}\large\frac{7}{9}+tan^{-1}\large\frac{1}{8}\)
Again by taking \(x=\frac{7}{9}\:and\:y=\frac{1}{8}\) in the same formula we get
\( = tan^{-1} \bigg[\large \frac{\large\frac{7}{9}+\large\frac{1}{8}}{1-\large\frac{7}{72}} \bigg] \)
\( = tan^{-1}1=\large\frac{\pi}{4}\)
=R.H.S.

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 22, 2013 by rvidyagovindarajan_1
 

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