1. Interpret the inverse trigonometric functions using principal value ranges

We use:

• $\sin^{-1} x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
• $\cos^{-1} x \in [0,\pi]$
• $\tan^{-1} x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

We evaluate each term separately.

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2. First term: $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$

Since

$$\sin \frac{2\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2},$$

we get

$$\sin^{-1}\left(\sin \frac{2\pi}{3}\right)=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3},$$

because $\frac{\pi}{3}\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

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3. Second term: $\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$

Now,

$$\cos \frac{7\pi}{6} = \cos \left(\pi + \frac{\pi}{6}\right) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}.$$

Hence,

$$\cos^{-1}\left(\cos \frac{7\pi}{6}\right)=\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right).$$

The principal value of $\cos^{-1}$ lies in $[0,\pi]$, and in this interval,

$$\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}.$$

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4. Third term: $\tan^{-1}\left(\tan \frac{3\pi}{4}\right)$

Since

$$\tan \frac{3\pi}{4} = -1,$$

we get

$$\tan^{-1}\left(\tan \frac{3\pi}{4}\right)=\tan^{-1}(-1)=-\frac{\pi}{4},$$

because the principal value range of $\tan^{-1}$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

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5. Add all three values

$$\frac{\pi}{3}+\frac{5\pi}{6}-\frac{\pi}{4}$$

Take LCM $=12$:

$$\frac{4\pi}{12}+\frac{10\pi}{12}-\frac{3\pi}{12} =\frac{11\pi}{12}.$$

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6. Match with options

$$\boxed{\frac{11\pi}{12}}$$

So the correct option is A.