1. **State the problem:** Prove that $$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}$$ for $x > 0$. 2. **Recall the formula for the tangent of a sum:** $$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$ 3. **Set** $a = \arctan x$ and $b = \arctan \frac{1}{x}$. 4. **Apply the tangent sum formula:** $$\tan(a + b) = \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} = \frac{x + \frac{1}{x}}{1 - 1} = \frac{x + \frac{1}{x}}{0}$$ 5. Since the denominator is zero, the tangent is undefined, which means the angle $a + b$ is an odd multiple of $\frac{\pi}{2}$. 6. Because both $a$ and $b$ are in the range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $x > 0$, their sum is exactly $\frac{\pi}{2}$. 7. **Therefore,** $$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}$$ This completes the proof.