1. **Problem statement:** Prove that $\arcsin x + \arccos x = 90^\circ$ or equivalently $\frac{\pi}{2}$ radians. 2. **Recall definitions:** - $\arcsin x$ is the angle $\theta$ such that $\sin \theta = x$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. - $\arccos x$ is the angle $\phi$ such that $\cos \phi = x$ and $\phi \in [0, \pi]$. 3. **Key identity:** For any angle $\theta$, $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$. 4. **Proof:** Let $\theta = \arcsin x$. Then by definition, $\sin \theta = x$. Using the identity, $\cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta = x$. Since $\arccos x$ is the angle whose cosine is $x$, and $\arccos x \in [0, \pi]$, it follows that: $$\arccos x = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \arcsin x$$ 5. **Rearranging:** $$\arcsin x + \arccos x = \arcsin x + \left(\frac{\pi}{2} - \arcsin x\right) = \frac{\pi}{2}$$ 6. **Conclusion:** Therefore, $\arcsin x + \arccos x = 90^\circ$ or $\frac{\pi}{2}$ radians, as required. This holds for all $x$ in the domain $[-1,1]$ where both inverse functions are defined.