The value of operatorname{cosec}^{-1}(sqrt{2} | vedclass

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Question

(A) We know the principal values of the inverse trigonometric functions:$\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$$\cos^{-1}\left(\frac{-1}

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(A) We know the principal values of the inverse trigonometric functions:$\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$$\cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$Substituting these values into the expression:$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$Finding a common denominator $(12)$:$\frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$

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

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