sin ^{-1}left[cos left(sin ^{-1} frac{sqrt{3} | vedclass

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Question

(B) We know that $\sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ because $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.Substituti

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(B) We know that $\sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ because $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.Substituting this into the expression,we get $\sin^{-1} \left[\cos \left(\frac{\pi}{3}\right)\right]$.Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,the expression becomes $\sin^{-1} \left(\frac{1}{2}\right)$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,therefore $\sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$.Thus,the correct option is $B$.

$\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]=$ . . . . . . .

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