Q.The principal value of $\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right)$ is

$\cot^{-1}$ has principal range $(0,\pi)$; the angle there with cotangent $-\tfrac{1}{\sqrt3}$ is $\tfrac{2\pi}{3}$.

Principal values of inverse trigonometric functions are a CBSE/NCERT Class 12 inverse trigonometric functions topic.

The principal-value range of $\cot^{-1}$ is $(0,\pi)$. We need $\theta$ in this interval with $\cot\theta = -\dfrac{1}{\sqrt3}$.

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Reduce $\sin(5\pi/6)$ to a simple value, then apply $\sin^{-1}$ within its principal range $[-\pi/2,\pi/2]$.

First compute $\sin(5\pi/6) = \sin(\pi - \pi/6) = \sin(\pi/6) = 1/2$.

Now we need $\sin^{-1}(1/2)$. The principal value branch of $\sin^{-1}$ is $[-\pi/2, \pi/2]$. Since $\pi/6$ lies in this range and $\sin(\pi/6) = 1/2$, we get $\sin^{-1}(\sin(5\pi/6)) = \pi/6$.

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