1. **State the problem:** We are given two equations involving inverse trigonometric functions: $$\sin^{-1}(\sqrt{x}) = \cos^{-1}(\sqrt{y})$$ and $$\tan\left(\tan^{-1}(3x) - \tan^{-1}(2y)\right) + \tan\left(\tan^{-1}(3y) - \tan^{-1}(2x)\right) = 1.$$ We need to find the values of $x$ and $y$. 2. **Use the identity relating inverse sine and inverse cosine:** Recall that for $\theta$ in $[0, \frac{\pi}{2}]$, $$\sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2}.$$ Since $\sin^{-1}(\sqrt{x}) = \cos^{-1}(\sqrt{y})$, this implies $$\sin^{-1}(\sqrt{x}) + \cos^{-1}(\sqrt{y}) = \frac{\pi}{2}.$$ Therefore, $$\sin^{-1}(\sqrt{x}) = \cos^{-1}(\sqrt{y}) = \theta,$$ and $$\theta + \theta = \frac{\pi}{2} \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}.$$ 3. **Find $x$ and $y$ from $\theta = \frac{\pi}{4}$:** Since $$\sin^{-1}(\sqrt{x}) = \frac{\pi}{4},$$ we have $$\sqrt{x} = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$$ Squaring both sides, $$x = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}.$$ Similarly, $$\cos^{-1}(\sqrt{y}) = \frac{\pi}{4} \implies \sqrt{y} = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$$ Squaring, $$y = \frac{1}{2}.$$ 4. **Check the second equation:** Recall the formula for tangent of difference: $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.$$ Let $$A = \tan^{-1}(3x), \quad B = \tan^{-1}(2y),$$ and $$C = \tan^{-1}(3y), \quad D = \tan^{-1}(2x).$$ Substitute $x = y = \frac{1}{2}$: $$3x = 3 \times \frac{1}{2} = \frac{3}{2}, \quad 2y = 2 \times \frac{1}{2} = 1,$$ $$3y = \frac{3}{2}, \quad 2x = 1.$$ Calculate each tangent difference: $$\tan(A - B) = \frac{\frac{3}{2} - 1}{1 + \frac{3}{2} \times 1} = \frac{\frac{1}{2}}{1 + \frac{3}{2}} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{5}.$$ Similarly, $$\tan(C - D) = \frac{\frac{3}{2} - 1}{1 + \frac{3}{2} \times 1} = \frac{1}{5}.$$ Sum these: $$\frac{1}{5} + \frac{1}{5} = \frac{2}{5} \neq 1.$$ 5. **Re-examine the second equation:** The sum of the two tangent expressions equals 1, but with $x = y = \frac{1}{2}$, it equals $\frac{2}{5}$. Try to find $x$ and $y$ such that the second equation holds. 6. **Use the tangent subtraction formula for both terms and sum:** $$\tan(\tan^{-1}(3x) - \tan^{-1}(2y)) = \frac{3x - 2y}{1 + 6xy},$$ $$\tan(\tan^{-1}(3y) - \tan^{-1}(2x)) = \frac{3y - 2x}{1 + 6xy}.$$ Sum: $$\frac{3x - 2y}{1 + 6xy} + \frac{3y - 2x}{1 + 6xy} = \frac{(3x - 2y) + (3y - 2x)}{1 + 6xy} = \frac{(3x - 2x) + (3y - 2y)}{1 + 6xy} = \frac{x + y}{1 + 6xy}.$$ Set equal to 1: $$\frac{x + y}{1 + 6xy} = 1 \implies x + y = 1 + 6xy.$$ 7. **Recall from step 3 that $x = y = \frac{1}{2}$ satisfies the first equation. Substitute into the above:** $$\frac{1}{2} + \frac{1}{2} = 1 + 6 \times \frac{1}{2} \times \frac{1}{2} \implies 1 = 1 + \frac{3}{2} = 2.5,$$ which is false. 8. **Solve the system:** From step 3, $\sqrt{x} = \sin \theta$, $\sqrt{y} = \cos \theta$, and $\theta = \frac{\pi}{4}$. From step 6, the second equation reduces to $$x + y = 1 + 6xy.$$ Substitute $y = 1 - x$ (from $x + y = 1 + 6xy$ rearranged): $$x + y = 1 + 6xy \implies y = 1 + 6xy - x.$$ But this is complicated; instead, use the relation from the first equation: $$\sqrt{x} = \sin \theta, \quad \sqrt{y} = \cos \theta,$$ so $$x = \sin^2 \theta, \quad y = \cos^2 \theta.$$ Substitute into the second equation: $$\sin^2 \theta + \cos^2 \theta = 1 + 6 \sin^2 \theta \cos^2 \theta.$$ Since $\sin^2 \theta + \cos^2 \theta = 1$, this becomes $$1 = 1 + 6 \sin^2 \theta \cos^2 \theta \implies 6 \sin^2 \theta \cos^2 \theta = 0.$$ 9. **Solve for $\theta$:** $$\sin^2 \theta \cos^2 \theta = 0 \implies \sin \theta = 0 \text{ or } \cos \theta = 0.$$ But $\theta$ must satisfy the first equation $\sin^{-1}(\sqrt{x}) = \cos^{-1}(\sqrt{y}) = \theta$, and $\theta$ is in $[0, \frac{\pi}{2}]$. If $\sin \theta = 0$, then $\theta = 0$, so $$x = \sin^2 0 = 0, \quad y = \cos^2 0 = 1.$$ If $\cos \theta = 0$, then $\theta = \frac{\pi}{2}$, so $$x = \sin^2 \frac{\pi}{2} = 1, \quad y = \cos^2 \frac{\pi}{2} = 0.$$ 10. **Check these solutions in the original equations:** - For $x=0$, $y=1$: $$\sin^{-1}(\sqrt{0}) = 0, \quad \cos^{-1}(\sqrt{1}) = \cos^{-1}(1) = 0,$$ so first equation holds. Second equation: $$\frac{x + y}{1 + 6xy} = \frac{0 + 1}{1 + 0} = 1,$$ which holds. - For $x=1$, $y=0$: $$\sin^{-1}(\sqrt{1}) = \sin^{-1}(1) = \frac{\pi}{2}, \quad \cos^{-1}(\sqrt{0}) = \cos^{-1}(0) = \frac{\pi}{2},$$ first equation holds. Second equation: $$\frac{1 + 0}{1 + 0} = 1,$$ which holds. **Final answers:** $$\boxed{(x,y) = (0,1) \text{ or } (1,0)}.$$