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Prove that $4\tan^{-1}\large\frac{1}{5}-\tan^{-1}\frac{1}{70}+\tan^{-1}\frac{1}{99}=\large\frac{\pi}{4}$

Can you answer this question?

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
edited Mar 19, 2013