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Let Sin⁻¹ x = A, Sin⁻¹ y = B

Sin A = x Sin B = y Cos A = √(1-x²) Cos B = √(1-y²)

Cos⁻¹ x = C => Cos C = x = Sin A = Cos (90-A) => C = 90 - A or 90 + A

Cos⁻¹ y = D => Cos D = y = SIn B => D = 90 + B or 90 - B

Now given equations are, A + B = 2π/3 -- equation 1 C+D = π/2 --- equation 2

1) C+D= (90 + A) + (90 + B) = π/2 A+B = -π/2 not possible

2) C+D = (90+A)+(90-B) = π/2 A-B = -π/2 -- equation 3 Hence, from equation 1 and 3, A = 15 deg. B = 105 deg

3) C+D = (90-A) + (90+B) = π/2 B-A = -π/2 , then B =15 deg, A = 105 deg

4) C+D = 90-A+90-B = π/2 A+B = π/2 this is not possible

x = Sin A = Sin 15 deg and y = Sin 105 deg or Cos 15 deg or, x = Cos 15 and y = Sin 15