Inverse Trigonometric Functions

Principal-value ranges are the central concept here, since every inverse trigonometric function is made single-valued by restricting its output to one specific branch, and ignoring those branches is the commonest source of error. The three functions involved each carry a different range, so they must be evaluated independently before summing. For ${\tan }^{-1}(1)$ the principal range is the open interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, and the angle whose tangent is one inside this range is $\frac{\pi }{4}$. For ${\cos }^{-1}\left(-\frac{1}{2}\right)$ the range is the closed interval $[0,\pi ]$, where the cosine equals $-\frac{1}{2}$ at $\frac{2\pi }{3}$. For ${\sin }^{-1}\left(-\frac{1}{2}\right)$ the range is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, and because the argument is negative the result is the negative angle $-\frac{\pi }{6}$. Adding with a common denominator of twelve gives $\frac{3\pi }{12}+\frac{8\pi }{12}-\frac{2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$. Option $\frac{5\pi }{6}$ arises from wrongly taking ${\sin }^{-1}\left(-\frac{1}{2}\right)$ as $+\frac{\pi }{6}$, ignoring the sign. Option $\frac{\pi }{2}$ drops the arctangent contribution. Option $\frac{7\pi }{12}$ mishandles the cosine branch by using $\frac{\pi }{3}$ instead of $\frac{2\pi }{3}$. This is the standard inverse-trigonometric principal-value evaluation pattern of JEE Advanced. As a final check, each computed angle indeed lies inside its own principal branch, so the summed value $\frac{3\pi }{4}$ is consistent and valid.