ISC Class 12 Mathematics — practice by topic

The principal-value range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. This is the unique interval on which $\sin$ is monotonic and covers its full range $[-1, 1]$. Quick reference: - $\sin^{-1}, \tan^{-1}, \csc^{-1}$: range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (with closed/open variations). - $\cos^{-1}, \sec^{-1}, \cot^{-1}$: range $[0, \pi]$ (option A is for these).

$\sin\theta = \dfrac{1}{2}$ has the principal solution $\theta = \dfrac{\pi}{6}$ (30°), which lies in the standard range $[-\pi/2, \pi/2]$. Memorise the three classic values: $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, $\sin^{-1}\dfrac{1}{\sqrt 2} = \dfrac{\pi}{4}$, $\sin^{-1}\dfrac{\sqrt 3}{2} = \dfrac{\pi}{3}$.

For $x \in [-1, 1]$: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$. Reason: $\cos(\pi - y) = -\cos y$. Setting $y = \cos^{-1}(x)$ gives $\cos(\pi - \cos^{-1} x) = -x$ so applying $\cos^{-1}$ to both sides (both LHS and target lie in $[0, \pi]$) gives the result. Compare $\sin^{-1}(-x) = -\sin^{-1}(x)$ (option A would apply here). The odd-function form works for $\sin^{-1}$ but not $\cos^{-1}$ because their principal ranges differ.