Prove that:

We know

$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \quad x y>1$

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

$=\tan ^{-1}\left(\frac{\frac{2\left(a b x^{2}-a b y^{2}+x y a^{2}-x y b^{2}\right)}{\left(a^{2}-b^{2}\right)\left(x^{2}-y^{2}\right)}}{\frac{\left(a^{2} x^{2}-a^{2} y^{2}-x^{2} b^{2}+y^{2} b^{2}-4 a b z y\right)}{\left(a^{2}-b^{2}\right)\left(x^{2}-y^{2}\right)}}\right)$

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

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

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