Question

For any a, b, x, y > 0, prove that:

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1  (2alphabeta)/(alpha^2-beta^2)`

`where  alpha =-ax+by, beta=bx+ay`

Answer 1

Let `a = btan m  and  x = ytan  n`

Then,

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=2/3tan^-1((3b^3tanm-b^3tan^3m)/(b^3-3b^3tan^2m))+2/3tan^-1((3y^3tann-y^3tan^3n)/(y^3-3y^3tan^2n))`

`=2/3tan^-1((3tanm-tan^3m)/(1-3tan^2m))+2/3tan^-1((3tann-tan^3n)/(1-3tan^2n))`

`=2/3tan^-1(tan3m)+2/3tan^-1(tan3n)`      `[because tan3x=(3tanx-tan^3x)/(1-3tan^2x)]`

`=2/3(3m)+2/3(3n)`

`=2m+2n`

`=2(tan^-1  a/b+tan^-1  x/y)`       `[because a=btanm, x=ytann]`

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

`=2tan^-1((ay+bx)/(by-ax))`

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

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

`=tan^-1{(2alphabeta)/(alpha^2-beta^2)}`      `[becausealpha=ay+bxandalpha=by-ax]`