$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) + \cos^{-1}\left(\cos\frac{7\pi}{6}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right)$$ is equal to

We need to evaluate:

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) + \cos^{-1}\left(\cos\frac{7\pi}{6}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right)$$

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

The range of $$\sin^{-1}$$ is $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$.

Since $$\frac{2\pi}{3}$$ is not in this range, we use: $$\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3}$$

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

Term 2: $$\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$$

The range of $$\cos^{-1}$$ is $$[0, \pi]$$.

Since $$\frac{7\pi}{6} > \pi$$, we use: $$\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$$

Since $$\frac{5\pi}{6} \in [0, \pi]$$:

$$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$$

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

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

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

$$\tan^{-1}(-1) = -\frac{\pi}{4}$$

Final Sum:

$$\frac{\pi}{3} + \frac{5\pi}{6} + \left(-\frac{\pi}{4}\right)$$

Taking LCM of 3, 6, and 4, which is 12:

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

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

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

The correct answer is Option A: $$\dfrac{11\pi}{12}$$.