Question

prove that

2tanx-1 = cos^-1 [ 1 - x² / 1 + x² ]

Answer 1

$2tan^{-1}x=cos^{-1}\frac{1-x^2}{1+x^2}$
$Let\:cos^{-1}\frac{1-x^2}{1+x^2}=A\\\\then,cosA=\frac{1-x^2}{1+x^2}\\\\so,tan\frac{A}{2}=\sqrt{\frac{1-cosA}{1+cosA}}$

$hence,tan\frac{A}{2}=\sqrt{\frac{1-\frac{1-x^2}{1+x^2}}{1+\frac{1-x^2}{1+x^2}}} \\ = \sqrt{ \frac{1 + {x}^{2} - 1 + {x}^{2} }{1 + {x}^{2} + 1 - {x}^{2} } } \\ = x$
now,
$tan\frac{A}{2}=x\\\\\frac{A}{2}=tan^{-1}x\\\\A=2tan^{-1}x\\\\cos^{-1}\frac{1-x^2}{1+x^2}=2tan^{-1}x$
hence proved //

Answer 2

ANSWER IS IN THE IMAGE