tan ^{-1}(-sqrt{3})-sec ^{-1}(-2)= . . . . . | vedclass

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Question

(C) We know that the principal value branch of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\sec^{-1} x$ is $[0, \pi] - \{\frac{\pi}{2}

Vedclass · Accepted Answer

$-\pi$

Vedclass · Answer

(C) We know that the principal value branch of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\sec^{-1} x$ is $[0, \pi] - \{\frac{\pi}{2}\}$.First,evaluate $	an^{-1}(-\sqrt{3})$:Since $	an(\frac{\pi}{3}) = \sqrt{3}$,we have $	an(-\frac{\pi}{3}) = -\sqrt{3}$.Thus,$	an^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.Next,evaluate $\sec^{-1}(-2)$:Since $\sec(\frac{\pi}{3}) = 2$,we have $\sec(\pi - \frac{\pi}{3}) = \sec(\frac{2\pi}{3}) = -2$.Thus,$\sec^{-1}(-2) = \frac{2\pi}{3}$.Finally,calculate the expression:$	an^{-1}(-\sqrt{3}) - \sec^{-1}(-2) = -\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{3\pi}{3} = -\pi$.Therefore,the correct option is $C$.

$\tan ^{-1}(-\sqrt{3})-\sec ^{-1}(-2)=$ . . . . . . .

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