Q.Find the principal value of $\cos^{-1}\!\left(-\dfrac{\sqrt3}{2}\right)$.

$\cot^{-1}$ 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 $\cot\theta = -\dfrac{1}{\sqrt3}$.

The cotangent is negative in the second quadrant. Since $\cot\dfrac{\pi}{3}=\dfrac{1}{\sqrt3}$, the corresponding second-quadrant angle is

$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3},$

and indeed $\cot\dfrac{2\pi}{3} = -\dfrac{1}{\sqrt3}$.

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 $\sin^{-1}(\sin(5\pi/6)) = \pi/6$.

Note that you cannot blindly cancel $\sin^{-1}\sin$ to get $5\pi/6$ back, because $5\pi/6$ lies outside the principal range of $\sin^{-1}$.