If x = (2tan^ - 1 3) and& y = ((1)/(2) ^ - 1 (4)/(3)),then

Let, $x = \sin 2\theta$ (where $tan \theta = 3$)
$⇒ x = \dfrac{2tan \theta}{1 + tan^{2}\theta} = \dfrac{6}{1 + 9} = \dfrac{3}{5}$
If $\alpha = tan^{- 1}\dfrac{4}{3}$

$⇒ y = \sin \left(\dfrac{\alpha}{2}\right) = \dfrac{1}{2}\sqrt{1 - cos\alpha}$
$= \dfrac{1}{\sqrt{2}}\sqrt{1 - \dfrac{3}{5}} = \dfrac{1}{\sqrt{5}}$
$∴ y^{2} = \dfrac{1}{5} ⇒ y^{2} = 2x - 1$