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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.`
Solution
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]`
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