If we are given the transformation:
$X=\ln (1+\sqrt x\cos y)$ $Y=2\sqrt x (1+\sqrt x\cos y)\sin y$
Would the inverse transformation be:
$x={Y^2\over 4\exp (2X)}+(\exp (X)-1)^2$
$y=\arctan \left({Y\over 2\exp(X)(\exp(X)-1)}\right)$ ?
If we are given the transformation:
$X=\ln (1+\sqrt x\cos y)$ $Y=2\sqrt x (1+\sqrt x\cos y)\sin y$
Would the inverse transformation be:
$x={Y^2\over 4\exp (2X)}+(\exp (X)-1)^2$
$y=\arctan \left({Y\over 2\exp(X)(\exp(X)-1)}\right)$ ?
$1+\sqrt x\cos y=e^X\implies \sqrt x\cos y=e^X-1-->(1)$
$Y=2\sqrt x(e^X)\sin y\implies \sqrt x\sin y=\frac Y{2e^X}-->(2)$
Squaring & adding we get, $x(\cos^2y+\sin^2y)=(e^X-1)^2+\left(\frac Y{2e^X}\right)^2$
or $x=(e^X-1)^2+\frac {Y^2}{4e^{2X}}$
Divide $(2)$ by $(1)$ (assuimg $x\ne 0$), $\tan y=\frac{\frac Y{2e^X}}{e^X-1}=\frac Y{e^{2X}(e^X-1)}$