A value of $$x$$ satisfying the equation $$\sin\left[\cot^{-1}(1 + x)\right] = \cos\left[\tan^{-1}x\right]$$, is:

Let $$\cot^{-1}(1 + x) = \theta$$. This means $$\cot \theta = \frac{1 + x}{1}$$.

In a right triangle with adjacent side 1+x and opposite side 1,

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

Therefore, $$\sin \theta = \frac{1}{\sqrt{x^2 + 2x + 2}}$$

Let $$\tan^{-1} x = \phi$$. This means $$\tan \phi = \frac{x}{1}$$.

In a right triangle where the opposite side is x and the adjacent side is 1,

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

Therefore, $$\cos \phi = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{1 + x^2}}$$

Now substitute the simplified forms back into the original equation:

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

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

$$1 + x^2 = x^2 + 2x + 2$$

$$1 = 2x + 2$$

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