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Question

If $$\displaystyle \tan ^{ -1 }{ y } =4\tan ^{ -1 }{ x } $$, then $$\displaystyle \frac { 1 }{ y } $$ is zero for


A
x=1±2
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B
x=2±3
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C
x=3±22
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D
all value of x
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Solution

The correct option is A $$x=1\pm \sqrt { 2 } $$
If we put $$\displaystyle x=\tan { \theta  } ,$$ the given equality becomes $$\displaystyle \tan ^{ -1 }{ y } =4\theta $$
$$\displaystyle \Rightarrow y=\tan { 4\theta  } =\frac { 2\tan { 2\theta  }  }{ 1-\tan ^{ 2 }{ 2\theta  }  } =\frac { 2\left[ \frac { 2\tan { \theta  }  }{ 1-\tan ^{ 2 }{ \theta  }  }  \right]  }{ 1-{ \left( \frac { 2\tan { \theta  }  }{ 1-\tan ^{ 2 }{ \theta  }  }  \right)  }^{ 2 } } $$
$$\displaystyle =\frac { 2\times 2x\left( 1-{ x }^{ 2 } \right)  }{ { \left( 1-{ x }^{ 2 } \right)  }^{ 2 }-4{ x }^{ 2 } } =\frac { 4x\left( 1-{ x }^{ 2 } \right)  }{ 1-6{ x }^{ 2 }+{ x }^{ 4 } } $$
$$\displaystyle \therefore \frac { 1 }{ y } $$ is zero if $$\displaystyle { x }^{ 4 }-6{ x }^{ 2 }+1=0$$
$$\displaystyle \Rightarrow { x }^{ 2 }=\frac { 6\pm \sqrt { 36-4 }  }{ 2 } =3\pm 2\sqrt { 2 } ={ \left( 1\pm \sqrt { 2 }  \right)  }^{ 2 }$$

Mathematics

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