If for some [0, ] , the relation x^2 + y^2 - 3 xy = 1 4 holds for all real numbers x, y (0,1) that satisfy ⁻¹x + ⁻¹y = , then the value of 3 times the sum of all possible values of is

Step-by-step solution

Let u = ⁻¹x and v = ⁻¹y . Then u + v = . Taking the sine of both sides: (u+v) = u v + u v = Since x, y (0,1) , we have u, v (0, 2 ) . Thus, u = x u = 1-x^2 and v = y v = 1-y^2 . Substituting these into the expansion: xy + 1-x^2 1-y^2 = 1-x^2 1-y^2 = - xy Squaring both sides: (1-x^2)(1-y^2) = ^2 - 2xy + x^2y^2 1 - x^2 - y^2 + x^2y^2 = ^2 - 2xy + x^2y^2 x^2 + y^2 - 2xy = 1 - ^2 = ^2 Comparing this with the given relation x^2 + y^2 - 3 xy = 1 4 , we get: 2 = 3 = 3 2 ^2 = 1 4 = 1 2 Since [0, ] , the possible values for