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Home  >>  CBSE XII  >>  Math
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Prove that $4\tan^{-1}\large\frac{1}{5}-\tan^{-1}\frac{1}{70}+\tan^{-1}\frac{1}{99}=\large\frac{\pi}{4}$

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Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}\), |x|<1
  • \( tan^{-1}x-tan^{-1}y=tan^{-1}\large\frac{x-y}{1+xy}\) , xy>-1
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy},\:\:\:xy<1 \)
L.H.S.
By taking \(x=\large\frac{1}{5},\:\large\frac{2x}{1-x^2}=\large\frac{2\frac{1}{5}}{1-\large\frac{1}{25}}=\large\frac{2}{5}.\large\frac{25}{24}=\large\frac{5}{12}\)
 
By substituting \(x=\large\frac{1}{5}\) in the above formula of \(2tan^{-1}x\) we get
\(\Rightarrow\: 2tan^{-1}\large\frac{1}{5}=tan^{-1}\large\frac{5}{12}\)
By taking \(x=\large\frac{5}{12},\frac{2x}{1-x^2}=\large\frac{2\frac{5}{12}}{1-\large\frac{25}{144}}=\large\frac{10}{12}.\large\frac{144}{119}=\large\frac{120}{119}\)
 
Apply once again the formula of \( 2tan^{-1}x\) by taking \( x=\large\frac{5}{12}\) we get
\(\Rightarrow\: 4tan^{-1}\large\frac{1}{5}=2tan^{-1}\large\frac{5}{12}=tan^{-1}\large\frac{120}{119}\)
By taking \(x=\large\frac{1}{99}\:and\:y=\large\frac{1}{70},\)
\(\large\frac{x-y}{1+xy}=\large\frac{\large\frac{1}{99}-\large\frac{1}{70}}{1+\large\frac{1}{99}.\large\frac{1}{70}}=-\large\frac{29}{6930}.\large\frac{6930}{6931}=-\large\frac{1}{239}\)
\(tan^{-1}\large\frac{1}{99}-tan^{-1}\large\frac{1}{70}=-tan^{-1}\large\frac{1}{239}\)
 
Substituting the values L.H.S. becomes,
\(\Rightarrow\: 4tan^{-1}\large\frac{1}{5}+(tan^{-1}\large\frac{1}{99}-tan^{-1}\large\frac{1}{70})=\)
\( tan^{-1}\large\frac{120}{119}-tan^{-1}\large\frac{1}{239} \)
 
Apply \( tan^{-1}x-tan^{-1}y\) formula by taking \(x=\large\frac{120}{119}\:and\:y=\large\frac{1}{239}\)
\( tan^{-1}\large\frac{120}{119}-tan^{-1}\large\frac{1}{239} =tan^{-1}\large\frac{\large\frac{120}{119}-\large\frac{1}{239}}{1-\large\large\frac{120}{119}.\large\frac{1}{239}}\)
\(=tan^{-1}1=\large\frac{\pi}{4}\)
=R.H.S.

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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