Question

for any a, b, x, y>0
prove that
2/3 tan-1(3ab^2-a^3/b^3-3a^2b)+ 2/3 tan-1(3xy^2-x^3/y^3-3x^2y) = tan-1 (2alpha*beta/ alpha ^2 - beta^2)

Answer 1

Well, let me use p for alpha and q for beta and I shall start for LHS. 
= arctan( 2pq/(p^2 - q^2) 
=arctan{2(q/p)/[1-(q/p)^2} 
=2arctan (q/p) 
=2arctan [(bx+ay)/(by-ax)] 
=2arctan{[(x/y) -(a/b)]/[1-(x/y)(a/b)]} 
=2[arctan(x/y) - arctan(a/b)] 
=(2/3)[3arctan(x/y) - 3arctan(a/b)] 
Now use 3arctan(m/n) = arctan{[3m/n) -(m/n)^3]/[1-3((m/n)^2] adn you get it = RHS.

Answer 2

(a) (x + 2)2 + (y + 1)2 = 0 x2 + 4x + 4 + y2 + 2y + 1 = 0 x2 + y2 + 4x + 2y + 5 = 0. (b) (x + 5) 2 + (y + 3)2 = 0 x2 + 10x + 25 + y2 + 6y + 9 = 0 .