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प्रश्न
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`
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उत्तर
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]`
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