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For any a,b,c,x,y >0 prove that : 2/3tan...

For any `a,b,c,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)/(a^2-beta^2)` where `alpha=-a x+b y ,beta=b x+a y`

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