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Find the value of $4\tan^{-1}{\large \frac{1}{5}}$$-\tan^{-1}{\large\frac{1}{239}}$

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  • \( 2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}\)
  • \( tan^{-1}x-tan^{-1}y=tan^{-1}\frac{x-y}{1+xy}\:\:xy<1 \)
  • \(tan^{-1}1=\frac{\pi}{4}\)
By taking \(x=\frac{1}{5},\:\frac{2x}{1-x^2}=\frac{2.\frac{1}{5}}{1-\frac{1}{25}}=\frac{2}{5}.\frac{25}{24}=\frac{5}{12}\)
Substituting in the above formula of \(2tan^{-1}x\) we get
\(2tan^{-1}\frac{1}{5}=tan^{-1}\frac{5}{12}\)
\( 4tan^{-1}\frac{1}{5} =2.(2tan^{-1}\frac{1}{5})= 2tan^{-1}\frac{5}{12} \)
Again by taking \(x=\frac{5}{12},\:\frac{2x}{1-x^2}=\frac{2.\frac{5}{12}}{1-\frac{25}{144}}=\frac{10}{12}.\frac{144}{119}=\frac{120}{119}\)
\( 2tan^{-1}\frac{5}{12}=tan^{-1}\frac{120}{119} \)
\( 4tan^{-1}\frac{1}{5}= 2tan^{-1}\frac{5}{12}=tan^{-1}\frac{120}{119} \)
Substituting the values, the given expression becomes
\( 4tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{239} \)
\( = tan^{-1}\frac{120}{119}-tan^{-1}\frac{1}{239}\)
By taking \(x=\frac{120}{119}\:and\:y=\frac{1}{239}\:we\:get\:\)
\(\frac{x-y}{1+xy}=\frac{\frac{120}{119}-\frac{1}{139}}{1+\frac{120}{119}.\frac{1}{239}}=1\)
Substituting in the above formula we get
\( tan^{-1}\frac{120}{119}-tan^{-1}\frac{1}{239}=tan^{-1}1=\frac{\pi}{4}\)
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answered Feb 19, 2013 by thanvigandhi_1
edited Mar 7, 2013 by rvidyagovindarajan_1
 
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