Question

Prove that $tan\{2tan^{-1}(\frac{\sqrt{1 + x^{2}} - 1}{x})\} = x$.

Answer 1

Given that

$LHS = tan[2tan^{-1}(\frac{\sqrt{1+x^{2}}-1}{x} )]$

We know that

$2tan^{-1}A= tan^{-1}\frac{2A}{1-A^{2} }$

so we have

$LHS = tan^{-1}\frac{2\frac{\sqrt{1+x^{2}}-1 }{x} }{1-(\frac{\sqrt{1+x^{2}}-1 }{x})^{2} }$

$LHS = tan[tan^{-1}[2\frac{\sqrt{1+x^{2}}-1 }{x} }]/[ {1-(\frac{\sqrt{1+x^{2}}-1 }{x})^{2} }]]\\\\LHS = tan[tan^{-1}[2\frac{\sqrt{1+x^{2}}-1 }{x} }][ \frac{x^{2} }{x^{2}-1-x^{2}-1+2\sqrt{1+x^{2} }}]] \\\\LHS = tan[tan^{-1}[2\frac{\sqrt{1+x^{2}}-1 }{x} }][ \frac{x^{2} }{-2+2\sqrt{1+x^{2} }}]]\\\\LHS=tan[tan^{-1}x]\\\\LHS=x=RHS$

Answer 2

Step by step solution:

As we know that

$2 {tan}^{ - 1} x = {tan}^{ - 1} (\frac{2x}{1 - {x}^{2} } )$
here
$= 2tan^{-1}(\frac{\sqrt{1 + x^{2}} - 1}{x}) \\ \\ \\ = tan^{-1}( \frac{ \frac{2 \sqrt{1 + {x}^{2} } - 1 }{x} }{1 - ( { \frac{ \sqrt{1 + {x} ^{2} } - 1}{x} })^{2} } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{(2 \sqrt{1 + {x}^{2} } - 1) x}{ {x}^{2} - 1 - {x}^{2} - 1 + 2 \sqrt{1 + {x}^{2} } } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{(2 \sqrt{1 + {x}^{2} } - 1)x}{ - 2 + 2 \sqrt{1 + {x}^{2} } } )$

${tan}^{ - 1}(x)$

so
$2tan^{-1}(\frac{\sqrt{1 + x^{2}} - 1}{x})=tan^{-1}x$

$tan[tan^{-1}x]=x\\\\=R.H.S.$
hence proved