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  4. Solve for x, tan -1(1/x)=π + tan -1x, 0 < x < 1

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Q. Solve for $ x,\,\,{{\tan }^{-1}}(1/x)=\pi +{{\tan }^{-1}}x,\,\,0 < x < 1 $

J & K CETJ & K CET 2014Inverse Trigonometric Functions

Solution:

We have, $ {{\tan }^{-1}}\left( \frac{1}{x} \right)=\pi +{{\tan }^{-1}}x,\,0 < x < 1 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{1}{x} \right)-{{\tan }^{-1}}x=\pi $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{\frac{1}{x}-x}{1+\frac{1}{x}.x} \right)=\pi $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{x(1+1)} \right)=\pi $
$ \Rightarrow $ $ \frac{1-{{x}^{2}}}{2x}=\tan \pi $
$ \Rightarrow $ $ \frac{1-{{x}^{2}}}{2x}=0 $
$ \Rightarrow $ $ 1-{{x}^{2}}=0\,\Rightarrow {{x}^{2}}=1\,\Rightarrow x=\pm 1 $ but given, $ 0 < x < 1 $