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

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(B) We know the following properties of inverse trigonometric functions: $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$,$\operatorname{cosec}^{-1}(-x) = -\opera

Vedclass · Accepted Answer

$\frac{-4}{5}$

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(B) We know the following properties of inverse trigonometric functions: $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$,$\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}(x)$,and $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.Also,$	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.$\operatorname{cosec}^{-1}(-\sqrt{2}) = -\operatorname{cosec}^{-1}(\sqrt{2}) = -\frac{\pi}{4}$.$\cos^{-1}\left(\frac{-1}{2}ight) = \pi - \cos^{-1}\left(\frac{1}{2}ight) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.Substituting these values into the expression:$\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} 	imes \frac{12}{5\pi} = -\frac{4}{5}$.

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$ is

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