We need to evaluate the expression: \[ \cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right) + \sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right) \] Step 1: Evaluate the first term \( \cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right) \) The cosine function is even, meaning \( \cos(-\theta) = \cos(\theta) \). \[ \cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) \] The principal range of \( \cos^{-1}(x) \) is \( [0, \pi] \). Since \( \frac{\pi}{6} \in [0, \pi] \), we have: \[ \cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} \] Step 2: Evaluate the second term \( \sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right) \) The principal range of \( \sin^{-1}(x) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). The angle \( \frac{5\pi}{6} \) is not in this range. We use the identity \( \sin(\pi - \theta) = \sin(\theta) \) to find an equivalent angle within the principal range: \[ \sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) \] Since \( \frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we get: \[ \sin^{-1}\left(\sin\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} \] Step 3: Add the two evaluated terms together: \[ \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \] Therefore, the value is \( \frac{\pi}{3} \).