Show that `(tan^-1)(2ab)/(a^2-b^2)+(tan^-1)(2xy)/(x^2-y^2)=(tan^-1) (2alphabeta)/(alpha^2-beta^2),` where `alpha=ax- by and beta=ay+bx.`

Explanation

`(tan^-1)(2ab)/(a^2-b^2)+(tan^-1)(2xy)/(x^2-y^2)`

`=tan^(-1)(frac{frac{2ab}{a^2-b^2}+frac{2xy}{x^2-y^2}}{1-frac{2ab}{a^2-b^2}xxfrac{2xy}{x^2-y^2}})`

`=tan^(-1)(frac{2abx^2-2aby^2+2xya^2-2xyb^2}{a^2x^2-x^2y^2-b^2x^2+b^2y^2-4abxy})`

`=tan^(-1) frac{2(bx+ay)(ax-by)}{(ax-by)^2-(ay+bx)^2}`

`=tan^(-1)frac{2 alpha beta}{alpha^2-beta^2}`