For any `a ,\ b ,\ x ,\ y >0` , prove that : `2/3tan^(-1)((3a b^2-a^3)/(b^3-3a^2b))+2/3tan^(-1)((3x y^2-x^3)/(y^3-3x^2y))=tan^(-1)(2alphabeta)/(alpha^2-beta^2)` , where `alpha=-a x+b y` `beta=b x+a ydot`

Explanation

`(2/3)tan^(-1)(frac{3(a/b)-(a^3/b^3)}{1-3(a^2/b^2)})+(2/3)tan^(-1)frac{3(x/y)-(x^3/y^3)}{1-3(x^2/y^2)}=tan^(-1) frac{2 alpha beta}{alpha^2-beta^2}`

`(2/3)xx 3tan^(-1)(a/b)+(2/3)tan^(-1)(x/y)=tan^(-1) frac{2 alpha beta}{alpha^2-beta^2}`

`2tan^(-1) frac{ay+bx}{by-ax}=tan^(-1)frac{2 (beta/alpha)}{1-(beta62/alpha^2)}`

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