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Prove that \[tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}= \frac{\pi}{4}\]

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  • \( tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1\)
Given $tan^{-1} \large\frac{1}{5}+ tan^{-1} \large\frac{1}{7}+tan^{-1}\large\frac{1}{3}+tan^{-1} \large\frac{1}{8}= \large\frac{\pi}{4}$
We will group two of the terms in the above L.H.S. equation and solve them using known inverse trignometric identities.
\(\Rightarrow L.H.S. = \bigg( tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}\bigg) + \bigg( tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8}\bigg) \)
Let's take the first term $tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}$
We know that \( tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1\)
By taking \(x=\large\frac{1}{5}\:and\:y=\large\frac{1}{7}\) in the above formula,
$x+y = \large\frac{1}{5}+\large\frac{1}{7} = \large\frac{12}{35}$
$1 - xy = 1 -\large \frac{1}{5} \times \frac{1}{7} = \large\frac{34}{35}$
$\large \frac{x+y}{1-xy}$ $= \large\frac{12}{35}.\large\frac{35}{34}=\large\frac{12}{34} = \large\frac{6}{17}$
\( \Rightarrow tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{7}=tan^{-1} \large\frac{6}{17}\)
Let's take the second term: $ tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8}$
Similarly by taking \(x=\large\frac{1}{3}\:and\:y=\large\frac{1}{8}\) we get:
$x+y = \large\frac{1}{3}+\large\frac{1}{8} = \large\frac{11}{24}$
$1-xy = 1-\large\frac{1}{3}\times \frac{1}{8} = \large\frac{23}{24}$
\(\large \frac{x+y}{1-xy}\) $ =\large\frac{11}{24}.\large\frac{24}{23}=\large\frac{11}{23}$
$\Rightarrow tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{8} = tan^{-1}\large\frac{11}{23}$
Therefore $tan^{-1} \large\frac{1}{5}+ tan^{-1} \large\frac{1}{7}+tan^{-1}\large\frac{1}{3}+tan^{-1}\large \frac{1}{8}$ now reduces to \( \Rightarrow\: tan^{-1} \large\frac{6}{17}+tan^{-1}\large\frac{11}{23}\)
Similarly by taking \(x=\large\frac{6}{17}\:and\:y=\large\frac{11}{23}\) we get:
$x+y = \large\frac{6}{17}+\large\frac{11}{23} = \large\frac{138+187}{391} =\large\frac{325}{391}$
$1-xy = 1 -\large \frac{6}{17} \times \frac{11}{23} = 1-\large\frac{66}{391} = =\large\frac{325}{391}$
\(\large \frac{x+y}{1-xy}=\large \frac{325}{391}\times \frac{391}{325}=1\)
\( \Rightarrow\) L.H.S. \(= tan^{-1} \large\frac{6}{17}+tan^{-1}\large\frac{11}{23} = tan^{-1}1\) \( = \large\frac{\pi}{4} \) = R.H.S.


answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
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