Question

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

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

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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