The principal value of cos ^{-1}left[frac{1}{ | vedclass

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(B) We need to find the principal value of $\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right]$.First,r

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$\frac{17 \pi}{20}$

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(B) We need to find the principal value of $\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}ight)ight]$.First,rewrite the expression inside the bracket:$\frac{1}{\sqrt{2}} \cos \frac{9 \pi}{10} - \frac{1}{\sqrt{2}} \sin \frac{9 \pi}{10}$Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have:$\cos \frac{\pi}{4} \cos \frac{9 \pi}{10} - \sin \frac{\pi}{4} \sin \frac{9 \pi}{10}$Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:$\cos \left(\frac{\pi}{4} + \frac{9 \pi}{10}ight) = \cos \left(\frac{5 \pi + 18 \pi}{20}ight) = \cos \left(\frac{23 \pi}{20}ight)$Since $\cos 	heta = \cos(2 \pi - 	heta)$,we have $\cos \left(\frac{23 \pi}{20}ight) = \cos \left(2 \pi - \frac{23 \pi}{20}ight) = \cos \left(\frac{17 \pi}{20}ight)$.Thus,$\cos ^{-1}\left[\cos \frac{17 \pi}{20}ight] = \frac{17 \pi}{20}$,which lies in the range $[0, \pi]$.Therefore,the correct option is $B$.

The principal value of $\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right]$ is

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