Question

Prove that:

`tan^-1  (2ab)/(a^2-b^2)+tan^-1  (2xy)/(x^2-y^2)=tan^-1  (2alphabeta)/(alpha^2-beta^2),`   where `alpha=ax-by  and  beta=ay+bx.`

Answer 1

We know

`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy)),xy>1`

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

`=tan^-1(((2(abx^2-aby^2+xya^2-xyb^2))/((a^2-b^2)(x^2-y^2)))/((a^2x^2-a^2y^2-x^2b^2+y^2b^2-4abxy)/((a^2-b^2)(x^2-y^2))))`

`=tan^-1((2(abx^2-aby^2+xya^2-xyb^2))/(a^2x^2-a^2y^2-x^2b^2+y^2b^2-2abxy))`

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

`=tan^-1((2alphabeta)/(alpha^2-beta^2))`    `[because alpha=ax-by  and  beta = ay+bx]`