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Question

Solve $$\displaystyle \tan^{-1} \left ( \frac{1-x}{1+x} \right ) = \frac{1}{2} \tan^{-1} x, \: \: \left( 0<x<1 \right) $$ 


Solution


$$\displaystyle \tan ^{ -1 } \left( \frac { 1-x }{ 1+x }  \right) =\frac { 1 }{ 2 } \tan ^{ -1 } x\Rightarrow 2\tan ^{ -1 } \left( \frac { 1-x }{ 1+x }  \right) =\tan ^{ -1 }{ x } $$
$$\displaystyle \Rightarrow \tan ^{ -1 }{ \left( \frac { 2\left( \frac { 1-x }{ 1+x }  \right)  }{ 1-{ \left( \frac { 1-x }{ 1+x }  \right)  }^{ 2 } }  \right)  } =\tan ^{ -1 }{ x } \Rightarrow \tan ^{ -1 }{ \left( \frac { 1-{ x }^{ 2 } }{ 2x }  \right)  } =\tan ^{ -1 }{ x } $$
$$\displaystyle \Rightarrow \frac { 1-{ x }^{ 2 } }{ 2x } =x\Rightarrow 1-{ x }^{ 2 }=2{ x }^{ 2 }\Rightarrow x=\pm \frac { 1 }{ \sqrt { 3 }  } $$
Neglecting $$\displaystyle x=-\frac { 1 }{ \sqrt { 3 }  } $$
We get $$\displaystyle x=\frac { 1 }{ \sqrt { 3 }  } $$

Mathematics

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