For any a, b, x, y > 0, prove that: 2 3 ^ -1 ( 3ab^2-a^3 b^3-3a^2b )+2 3 ^ -1 ( 3xy^2-x^3 y^3-3x^2y )= ^ -1 2 ^2- ^2 where =-ax+by, =bx+ay

Let a=b and x=y Then 2 3 ^ -1 ( 3ab^2-a^3 b^3-3a^2b )+2 3 ^ -1 ( 3xy^2-x^3 y^3-3x^2y ) =2 3 ^ -1 ( 3b^3 -b^3 ^3m b^3-3b^3 ^2m )+2 3 ^ -1 ( 3y^3 -y^3 ^3n y^3-3y^3 ^2n ) =2 3 ^ -1 ( 3 - ^3m 1-3 ^2m )+2 3 ^ -1 ( 3 - ^3n 1- ^2n ) =2 3 ^ -1 ( 3m)+2 3 ^ -1 ( 3n) [ a=b ,x=y= ] =2 ^ -1 ( a b + x y 1- a b x y ) =2 ^ -1 ( ay+bx by-ax ) = ^ -1 2 ay+bx by-ax 1- ( ay+bx by-ax )^2 = ^ -1 2(ay+bx)(by-ax) (by-ax)^2-(ay+bx)^2 = ^ -1 2 ^2- ^2 [ =bx+ay, =-ax+by]

For any a, b, x, y > 0, prove that: 2 3 ^ -1 ( 3ab^2-a^3 b^3-3a^2…

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Question

For any a, b, x, y > 0, prove that:
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$
where $\alpha=-\text{ax}+\text{by},\beta=\text{bx}+\text{ay}$

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Answer

Let $\text{a}=\text{b}\tan\text{m}$ and $\text{x}=\text{y}\tan\text{n}$
Then
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\text{b}^3\tan\text{m}-\text{b}^3\tan^3\text{m}}{\text{b}^3-3\text{b}^3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{y}^3\tan\text{n}-\text{y}^3\tan^3\text{n}}{\text{y}^3-3\text{y}^3\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{m}-\tan^3\text{m}}{1-3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{n}-\tan^3\text{n}}{1-\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}(\tan3\text{m})+\frac{2}{3}\tan^{-1}(\tan3\text{n})$ $[\because\ \text{a}=\text{b}\tan\text{m},\text{x}=\text{y}=\tan\text{n}]$
$=2\tan^{-1}\Bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{x}}{\text{y}}}{1-\frac{\text{a}}{\text{b}}\frac{\text{x}}{\text{y}}}\Bigg)$
$=2\tan^{-1}\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)$
$=\tan^{-1}\begin{Bmatrix}\frac{2\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}}{1-\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)^2}\end{Bmatrix}$
$=\tan^{-1}\Bigg\{\frac{2(\text{ay}+\text{bx})(\text{by}-\text{ax})}{(\text{by}-\text{ax})^2-(\text{ay}+\text{bx})^2}\Bigg\}$
$=\tan^{-1}\Big\{\frac{2\alpha\beta}{\alpha^2-\beta^2}\Big\}$ $[\because\ \beta=\text{bx}+\text{ay},\alpha=-\text{ax}+\text{by}]$

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