If x = √(asin⁻¹t), y = √(acos⁻¹t), show that dy/dx = -y/x

x = √(asin⁻¹t), y = √(acos⁻¹t) ⇒ x² = asin⁻¹t, y² = acos⁻¹t
dy/dx = dy/dt × dt/dx
Now,
2y dy/dt = acos⁻¹t × lna × ⁻¹/√(1-t²) = y² × lna × ⁻¹/√(1-t²) .. (1)
2x dx/dt = asin⁻¹t × lna × 1/√(1-t²) = x² × lna × 1/√(1-t²) .. (2)
Dividing equations (1) and (2), we have
dy/dx × y/x = -y²/x² ⇒ dy/dx = -y/x