Question:

To evaluate the expression $\tan^{-1}\left(\tan\dfrac{5\pi}{6}\right)+\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)$, each term is evaluated separately. First, consider the term $\tan^{-1}\left(\tan\dfrac{5\pi}{6}\right)$. The principal value branch for $\tan^{-1}(x)$ is $(-\pi/2, \pi/2)$. Since $\dfrac{5\pi}{6}$ is not within this interval, we use the identity $\tan(\theta) = \tan(\theta - n\pi)$ for an integer $n$. $$ \tan\left(\dfrac{5\pi}{6}\right) = \tan\left(\dfrac{5\pi}{6} - \pi\right) = \tan\left(-\dfrac{\pi}{6}\right) $$ Since $-\dfrac{\pi}{6} \in (-\pi/2, \pi/2)$, $$ \tan^{-1}\left(\tan\dfrac{5\pi}{6}\right) = \tan^{-1}\left(\tan\left(-\dfrac{\pi}{6}\right)\right) = -\dfrac{\pi}{6} $$ Next, consider the term $\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)$. The principal value branch for $\cos^{-1}(x)$ is $[0, \pi]$. Since $\dfrac{13\pi}{6}$ is not within this interval, we use the identity $\cos(\theta) = \cos(\theta - 2n\pi)$ for an integer $n$. $$ \cos\left(\dfrac{13\pi}{6}\right) = \cos\left(\dfrac{13\pi}{6} - 2\pi\right) = \cos\left(\dfrac{13\pi - 12\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) $$ Since $\dfrac{\pi}{6} \in [0, \pi]$, $$ \cos^{-1}\left(\cos\dfrac{13\pi}{6}\right) = \cos^{-1}\left(\cos\left(\dfrac{\pi}{6}\right)\right) = \dfrac{\pi}{6} $$ Finally, sum the two evaluated terms: $$ \tan^{-1}\left(\tan\dfrac{5\pi}{6}\right)+\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right) = -\dfrac{\pi}{6} + \dfrac{\pi}{6} = 0 $$