Find the value of tan ^{-1}(1)+cos ^{-1}left( | vedclass

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(A) Let $	an ^{-1}(1)=x$. Then,$	an x=1=	an \left(\frac{\pi}{4}ight)$. Thus,$	an ^{-1}(1)=\frac{\pi}{4}$.Let $\cos ^{-1}\left(-\frac{1}{2}ight

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$\frac{3 \pi}{4}$

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(A) Let $	an ^{-1}(1)=x$. Then,$	an x=1=	an \left(\frac{\pi}{4}ight)$. Thus,$	an ^{-1}(1)=\frac{\pi}{4}$.Let $\cos ^{-1}\left(-\frac{1}{2}ight)=y$. Then,$\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}ight)=\cos \left(\pi-\frac{\pi}{3}ight)=\cos \left(\frac{2 \pi}{3}ight)$. Thus,$\cos ^{-1}\left(-\frac{1}{2}ight)=\frac{2 \pi}{3}$.Let $\sin ^{-1}\left(-\frac{1}{2}ight)=z$. Then,$\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}ight)=\sin \left(-\frac{\pi}{6}ight)$. Thus,$\sin ^{-1}\left(-\frac{1}{2}ight)=-\frac{\pi}{6}$.Substituting these values,we get:$	an ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}ight)+\sin ^{-1}\left(-\frac{1}{2}ight) = \frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$.Taking the least common multiple of the denominators $4, 3, 6$,which is $12$:$= \frac{3 \pi + 8 \pi - 2 \pi}{12} = \frac{9 \pi}{12} = \frac{3 \pi}{4}$.

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

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