tan ^{-1} sqrt{3} - sec ^{-1}(-2) is equal to | vedclass

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(D) Let $	an ^{-1} \sqrt{3} = x$. Then,$	an x = \sqrt{3} = 	an \frac{\pi}{3}$. We know that the range of the principal value branch of $	an ^{-1}$

Vedclass · Accepted Answer

$-\frac{\pi}{3}$

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(D) Let $	an ^{-1} \sqrt{3} = x$. Then,$	an x = \sqrt{3} = 	an \frac{\pi}{3}$. We know that the range of the principal value branch of $	an ^{-1}$ is $\left( -\frac{\pi}{2}, \frac{\pi}{2} ight)$. Therefore,$	an ^{-1} \sqrt{3} = \frac{\pi}{3}$. Let $\sec ^{-1}(-2) = y$. Then,$\sec y = -2 = -\sec \left( \frac{\pi}{3} ight) = \sec \left( \pi - \frac{\pi}{3} ight) = \sec \frac{2 \pi}{3}$. We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi] - \left\{ \frac{\pi}{2} ight\}$. Therefore,$\sec ^{-1}(-2) = \frac{2 \pi}{3}$. Thus,$	an ^{-1}(\sqrt{3}) - \sec ^{-1}(-2) = \frac{\pi}{3} - \frac{2 \pi}{3} = -\frac{\pi}{3}$.

$\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2)$ is equal to

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