The value of frac{tan^{-1}(1) + cos^{-1}(frac | vedclass

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(D) Step $1$: Evaluate each inverse trigonometric term.$	an^{-1}(1) = \frac{\pi}{4}$$\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$$\sin^{-1}(\frac{1}{2}) =

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$-\frac{9}{4}$

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(D) Step $1$: Evaluate each inverse trigonometric term.$	an^{-1}(1) = \frac{\pi}{4}$$\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$$\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$$	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$Step $2$: Substitute these values into the expression.Numerator: $\frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{6} = \frac{3\pi + 4\pi + 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$Denominator: $\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$Step $3$: Calculate the final ratio.$\frac{3\pi/4}{-\pi/3} = \frac{3\pi}{4} 	imes (-\frac{3}{\pi}) = -\frac{9}{4}$

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

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