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

Toolbox:
• $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$
• $tan^{-1}x-tan^{-1}y=tan^{-1}\frac{x-y}{1+xy}\:\:xy<1$
• $tan^{-1}1=\frac{\pi}{4}$
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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edited Mar 7, 2013