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# Prove that : $tan^{-1}\bigg(\frac{1}{3} \bigg)+tan^{-1}\bigg(\frac{1}{5}\bigg)+tan^{-1}\bigg(\frac{1}{7}\bigg)+tan^{-1}\bigg(\frac{1}{8}\bigg)=\large\frac{\pi}{4}$

This question is Q.No. 8 of misc. of chapter 2

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## 1 Answer

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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 Mar 1, 2013
edited Mar 17, 2013

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