If $$\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1 - x^2})$$, $$x \in (0, 1)$$, then the value of $$x$$ is :

Let $$\theta = \tan^{-1}(x\sqrt{2}) \implies \tan\theta = \frac{x\sqrt{2}}{1} = \frac{\text{Perpendicular}}{\text{Base}}$$

$$\text{Hypotenuse} = \sqrt{(x\sqrt{2})^2 + 1^2} = \sqrt{2x^2 + 1}$$

$$\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{x\sqrt{2}}{\sqrt{2x^2 + 1}} \implies \theta = \sin^{-1}\left(\frac{x\sqrt{2}}{\sqrt{2x^2 + 1}}\right)$$

$$\text{LHS} = \sin\left(\sin^{-1}\left(\frac{x\sqrt{2}}{\sqrt{2x^2 + 1}}\right)\right) = \frac{x\sqrt{2}}{\sqrt{2x^2 + 1}}$$

Let $$\phi = \sin^{-1}\left(\sqrt{1 - x^2}\right) \implies \sin\phi = \frac{\sqrt{1 - x^2}}{1} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$$

$$\text{Base} = \sqrt{1^2 - \left(\sqrt{1 - x^2}\right)^2} = \sqrt{1 - (1 - x^2)} = \sqrt{x^2} = x \quad (\text{since } x \in (0,1))$$

$$\cot\phi = \frac{\text{Base}}{\text{Perpendicular}} = \frac{x}{\sqrt{1 - x^2}} \implies \phi = \cot^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right)$$

$$\text{RHS} = \cot\left(\cot^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right)\right) = \frac{x}{\sqrt{1 - x^2}}$$

$$\frac{x\sqrt{2}}{\sqrt{2x^2 + 1}} = \frac{x}{\sqrt{1 - x^2}}$$

$$\frac{\sqrt{2}}{\sqrt{2x^2 + 1}} = \frac{1}{\sqrt{1 - x^2}}$$ ($$x \neq 0$$)

$$\frac{2}{2x^2 + 1} = \frac{1}{1 - x^2}$$

$$2(1 - x^2) = 1(2x^2 + 1)$$

$$1 = 4x^2 \implies x^2 = \frac{1}{4}$$

$$x = \frac{1}{2}$$