Let the line y - x = 1 intersect the ellipse $$\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is:

Substituting $$y = x + 1$$ into the ellipse equation gives

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

which simplifies to

$$\frac{x^2}{2} + x^2 + 2x + 1 = 1 \quad\Longrightarrow\quad \frac{3x^2}{2} + 2x = 0 \quad\Longrightarrow\quad x(3x + 4) = 0,$$

so that $$x = 0$$ or $$x = -\frac{4}{3}\,.$$

When $$x = 0$$, we have $$y = 1$$, hence $$A = (0,1)$$. When $$x = -\frac{4}{3}$$, we get $$y = -\frac{4}{3} + 1 = -\frac{1}{3}$$, so $$B = \bigl(-\tfrac{4}{3}, -\tfrac{1}{3}\bigr)\,.$$

Next, we denote $$\overrightarrow{OA} = (0,1)$$ and $$\overrightarrow{OB} = \bigl(-\tfrac{4}{3}, -\tfrac{1}{3}\bigr)$$ and use the dot-product formula

$$\cos\theta = \frac{\overrightarrow{OA}\cdot\overrightarrow{OB}}{\lvert\overrightarrow{OA}\rvert\,\lvert\overrightarrow{OB}\rvert}\,.$$

Since $$\overrightarrow{OA}\cdot\overrightarrow{OB} = 0\cdot\bigl(-\tfrac{4}{3}\bigr) + 1\cdot\bigl(-\tfrac{1}{3}\bigr) = -\frac{1}{3}$$,

$$\lvert\overrightarrow{OA}\rvert = 1,\qquad \lvert\overrightarrow{OB}\rvert = \sqrt{\tfrac{16}{9}+\tfrac{1}{9}} = \frac{\sqrt{17}}{3},$$

$$\cos\theta = \frac{-\tfrac{1}{3}}{1\cdot(\sqrt{17}/3)} = -\frac{1}{\sqrt{17}}\,.$$

Because $$\cos\theta = -\frac{1}{\sqrt{17}}$$, it follows that

$$\theta = \pi - \cos^{-1}\!\bigl(\tfrac{1}{\sqrt{17}}\bigr)\,.$$

On the other hand, if we set $$\tan\phi = 4$$ with $$\phi = \tan^{-1}(4)$$, then in a right triangle with opposite side 4 and adjacent side 1 the hypotenuse is $$\sqrt{17}$$, so $$\cos\phi = \tfrac{1}{\sqrt{17}}$$ and hence $$\cos^{-1}\!\bigl(\tfrac{1}{\sqrt{17}}\bigr) = \tan^{-1}(4)\,.$$

Moreover, by the complementary-angle identity, $$\tan^{-1}(4) = \tfrac{\pi}{2} - \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr)\,.$$ Therefore

$$\theta = \pi - \Bigl(\tfrac{\pi}{2} - \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr)\Bigr) = \frac{\pi}{2} + \tan^{-1}\!\bigl(\tfrac{1}{4}\bigr)\,.$$