frac{tan ^{-1}(sqrt{3})-sec ^{-1}(-2)}{operat | vedclass

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Question

(B) We know the principal values of the inverse trigonometric functions: $	an ^{-1}(\sqrt{3}) = \frac{\pi}{3}$ $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2)

Vedclass · Accepted Answer

$-\frac{4}{5}$

Vedclass · Answer

(B) We know the principal values of the inverse trigonometric functions: $	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: $\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} 	imes \frac{12}{5\pi} = -\frac{4}{5}$

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

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