The value of tan ^{-1}(1)+cos ^{-1}left(-frac | vedclass

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(D) We know that $	an ^{-1}(1) = \frac{\pi}{4}$.We also know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,so $\cos ^{-1}\left(-\frac{1}{2}ight) = \p

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

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(D) We know that $	an ^{-1}(1) = \frac{\pi}{4}$.We also know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,so $\cos ^{-1}\left(-\frac{1}{2}ight) = \pi - \cos ^{-1}\left(\frac{1}{2}ight) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.We know that $\sin ^{-1}(-x) = -\sin ^{-1}(x)$,so $\sin ^{-1}\left(-\frac{1}{2}ight) = -\sin ^{-1}\left(\frac{1}{2}ight) = -\frac{\pi}{6}$.Adding these values together:$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.

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

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