The value of tan^{-1}(sqrt{3}) + sec^{-1}(-2) | vedclass

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(C) Step $1$: Evaluate $	an^{-1}(\sqrt{3})$. Since $	an(\frac{\pi}{3}) = \sqrt{3}$, $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Step $2$: Evaluate $\sec^{

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$\frac{7\pi}{6}$

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(C) Step $1$: Evaluate $	an^{-1}(\sqrt{3})$. Since $	an(\frac{\pi}{3}) = \sqrt{3}$, $	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Step $2$: Evaluate $\sec^{-1}(-2)$. Since $\sec(\frac{2\pi}{3}) = -2$, $\sec^{-1}(-2) = \frac{2\pi}{3}$.Step $3$: Evaluate $\sin^{-1}(-\frac{1}{2})$. Since $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$, $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$.Step $4$: Substitute the values into the expression: $\frac{\pi}{3} + \frac{2\pi}{3} - (-\frac{\pi}{6}) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}(-\frac{1}{2})$ is

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The correct answer is:

$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=$

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