# For any a, b, x, y > 0, prove that:

Question:

For any abxy > 0, prove that:

$\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1} \frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}$

where α = − ax + by, β = bx + ay

Solution:

Let $a=b \tan m$ and $x=y \tan n$

Then,

$\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\frac{2}{3} \tan ^{-1}\left(\frac{3 b^{3} \tan m-b^{3} \tan ^{3} m}{b^{3}-3 b^{3} \tan ^{2} m}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 y^{3} \tan n-y^{3} \tan ^{3} n}{y^{3}-3 y^{3} \tan ^{2} n}\right)$

$=\frac{2}{3} \tan ^{-1}\left(\frac{3 \tan m-\tan ^{3} m}{1-3 \tan ^{2} m}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 \tan n-\tan ^{3} n}{1-\tan ^{2} n}\right)$

$=\frac{2}{3} \tan ^{-1}(\tan 3 m)+\frac{2}{3} \tan ^{-1}(\tan 3 n)$                  $\left[\because \tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\right]$

$=\frac{2}{3}(3 m)+\frac{2}{3}(3 n)$

$=2 m+2 n$

$=2\left(\tan ^{-1} \frac{a}{b}+\tan ^{-1} \frac{x}{y}\right) \quad[\because a=b \tan m, x=y \tan n]$

$=2 \tan ^{-1}\left(\frac{\frac{a}{b}+\frac{x}{y}}{1-\frac{a}{b} \frac{x}{y}}\right)$

$=2 \tan ^{-1}\left(\frac{a y+b x}{b y-a x}\right)$

$=\tan ^{-1}\left\{\frac{2 \frac{a y+b x}{b y-a x}}{1-\left(\frac{a y+b x}{b y-a x}\right)^{2}}\right\}$

$=\tan ^{-1}\left\{\frac{2(a y+b x)(b y-a x)}{(b y-a x)^{2}-(a y+b x)^{2}}\right\}$

$=\tan ^{-1}\left\{\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right\}$       $[\because \beta=a y+b x$ and $\alpha=b y-a x]$