The value of $sin^{-1} \left[cos \begin{Bmatrix}sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\end{Bmatrix}\right]$, is
Answer & explanation
Correct answer: option 2
$sin^{-1} \left[cos \begin{Bmatrix}sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\end{Bmatrix}\right]$
$sin^{-1}\begin{Bmatrix}cos \left(-\frac{\pi}{3}\right)\end {Bmatrix}$ $\left[∵sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}\right]$
$= sin^{-1}\left(cos\frac{\pi}{3}\right) - sin^{-1} (\frac{1}{2}) =\frac{\pi}{6}$