$-\dfrac{2 \pi}{5}$ is the principal value of

The principal value branch of $\sin^{-1}(x)$ is $[-\pi/2, \pi/2]$. For $\sin^{-1}(\sin(7\pi/5))$, we write $7\pi/5 = \pi + 2\pi/5$. Thus, $\sin(7\pi/5) = \sin(\pi + 2\pi/5) = -\sin(2\pi/5) = \sin(-2\pi/5)$. Since $-2\pi/5 \in [-\pi/2, \pi/2]$, the value is $-2\pi/5$.

The principal value branch of $\sec^{-1}(x)$ is $[0, \pi] \setminus \{\pi/2\}$. For $\sec^{-1}(\sec(7\pi/5))$, we have $7\pi/5 = 2\pi - 3\pi/5$, so $\sec(7\pi/5) = \sec(3\pi/5)$. Since $3\pi/5 \in [0, \pi]$, the value is $3\pi/5$.

The principal value branch of $\tan^{-1}(x)$ is $(-\pi/2, \pi/2)$. For $\tan^{-1}(\tan(7\pi/5))$, we have $7\pi/5 = \pi + 2\pi/5$, so $\tan(7\pi/5) = \tan(2\pi/5)$. Since $2\pi/5 \in (-\pi/2, \pi/2)$, the value is $2\pi/5$.

The principal value branch of $\cos^{-1}(x)$ is $[0, \pi]$. For $\cos^{-1}(\cos(7\pi/5))$, we have $7\pi/5 = 2\pi - 3\pi/5$, so $\cos(7\pi/5) = \cos(3\pi/5)$. Since $3\pi/5 \in [0, \pi]$, the value is $3\pi/5$.

Comparing the results, only $\sin^{-1}(\sin(7\pi/5))$ yields $-2\pi/5$.

Answer: $\sin ^{-1}\left[\sin \left(\dfrac{7 \pi}{5}\right)\right]$