The value of tan ^{-1}(-1)+sec ^{-1}(-2)+sin | vedclass

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(D) We know the principal value branches of the inverse trigonometric functions:$1$. $	an ^{-1}(-x) = -	an ^{-1}(x)$,so $	an ^{-1}(-1) = -\frac{\pi

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$\frac{2 \pi}{3}$

Vedclass · Answer

(D) We know the principal value branches of the inverse trigonometric functions:$1$. $	an ^{-1}(-x) = -	an ^{-1}(x)$,so $	an ^{-1}(-1) = -\frac{\pi}{4}$.$2$. $\sec ^{-1}(-x) = \pi - \sec ^{-1}(x)$,so $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.$3$. $\sin ^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.Adding these values together:$-\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{4} = \frac{2\pi}{3}$.

The value of $\tan ^{-1}(-1)+\sec ^{-1}(-2)+\sin ^{-1} \frac{1}{\sqrt{2}}$ is . . . . . . .

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