sin ^{-1}left(cos frac{pi}{13}right)+cos ^{-1 | vedclass

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(C) We know that $\cos 	heta = \sin \left(\frac{\pi}{2} - 	hetaight)$ and $\sin 	heta = \cos \left(\frac{\pi}{2} - 	hetaight)$.Substituting th

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$\frac{11 \pi}{13}$

Vedclass · Answer

(C) We know that $\cos 	heta = \sin \left(\frac{\pi}{2} - 	hetaight)$ and $\sin 	heta = \cos \left(\frac{\pi}{2} - 	hetaight)$.Substituting these into the expression:$\sin ^{-1}\left(\sin \left(\frac{\pi}{2} - \frac{\pi}{13}ight)ight) + \cos ^{-1}\left(\cos \left(\frac{\pi}{2} - \frac{\pi}{13}ight)ight)$Since $\frac{\pi}{2} - \frac{\pi}{13} = \frac{11\pi}{26}$,which lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$ and $[0, \pi]$ for $\cos^{-1}$,we have:$= \left(\frac{\pi}{2} - \frac{\pi}{13}ight) + \left(\frac{\pi}{2} - \frac{\pi}{13}ight)$$= \pi - \frac{2\pi}{13}$$= \frac{13\pi - 2\pi}{13} = \frac{11\pi}{13}$.

$\sin ^{-1}\left(\cos \frac{\pi}{13}\right)+\cos ^{-1}\left(\sin \frac{\pi}{13}\right) = $ . . . . . . .

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