The value of $$\tan^{-1}\left[\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right]$$, $$|x| \lt \frac{1}{2}$$, $$x \neq 0$$, is equal to:

We have to evaluate the expression

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$$

where the domain condition $$|x|\lt \dfrac12,\;x\neq0$$ is given so that every square-root and inverse-trig quantity involved is real and single-valued in its principal branch.

First we recall the tangent addition formula written in the form

$$\tan\!\left(\frac{\pi}{4}+y\right)=\frac{1+\tan y}{1-\tan y}.$$

Comparing this standard result with the fraction that appears inside the inverse tangent, we try to bring that fraction to the pattern $$\dfrac{1+\tan y}{1-\tan y}.$$

Let us denote

$$A=\sqrt{1+x^2},\qquad B=\sqrt{1-x^2}.$$

Then the argument of $$\tan^{-1}$$ is

$$\frac{A+B}{A-B}.$$

We divide the numerator and the denominator by $$A$$ (which is positive, because $$1+x^2\gt 0$$):

$$\frac{A+B}{A-B}=\frac{\dfrac{A}{A}+\dfrac{B}{A}}{\dfrac{A}{A}-\dfrac{B}{A}}=\frac{1+\dfrac{B}{A}}{1-\dfrac{B}{A}}.$$

Now set

$$\tan y=\frac{B}{A}=\frac{\sqrt{1-x^2}}{\sqrt{1+x^2}}.$$

Because $$0\lt |x|\lt \dfrac12$$, we have $$0\lt \frac{B}{A}\lt 1$$, so $$y$$ lies in the interval $$\left(0,\dfrac{\pi}{4}\right),$$ well inside the principal branch. With this choice we obtain

$$\frac{A+B}{A-B}=\frac{1+\tan y}{1-\tan y}=\tan\!\left(\frac{\pi}{4}+y\right).$$

Therefore

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]=\frac{\pi}{4}+y.$$

Our task is now reduced to expressing $$y$$ in terms of $$x$$. We already know

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

To convert this to a cosine-inverse expression we use the double-angle identity

$$\cos 2y=\frac{1-\tan^2 y}{1+\tan^2 y}.$$

Compute the square of the tangent:

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

Now substitute in the identity:

$$$ \begin{aligned} \cos 2y &=\frac{1-\dfrac{1-x^2}{1+x^2}}{1+\dfrac{1-x^2}{1+x^2}} =\frac{\dfrac{(1+x^2)-(1-x^2)}{1+x^2}}{\dfrac{(1+x^2)+(1-x^2)}{1+x^2}} =\frac{2x^2}{2} =x^2. \end{aligned} $$$

We have obtained $$\cos 2y=x^2.$$ The principal value of $$\cos^{-1}$$ lies in $$[0,\pi],$$ and because $$y\in\!\left(0,\dfrac{\pi}{4}\right)$$ we get $$2y\in\!\left(0,\dfrac{\pi}{2}\right)\subset[0,\pi],$$ so no ambiguity arises. Hence

$$2y=\cos^{-1}(x^2)\quad\Longrightarrow\quad y=\frac12\,\cos^{-1}(x^2).$$

Substituting this back into the earlier result gives

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]=\frac{\pi}{4}+\frac12\,\cos^{-1}(x^2).$$

Thus the given expression equals $$\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}x^2,$$ which exactly matches Option A.

Hence, the correct answer is Option A.