cos ^{-1}left(frac{-1}{2}right)-2 sin ^{-1}le | vedclass

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(D) We use the properties of inverse trigonometric functions: $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,$\sin ^{-1}(-x) = -\sin ^{-1}(x)$,and $	an ^{-1}

Vedclass · Accepted Answer

$\frac{43 \pi}{12}$

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(D) We use the properties of inverse trigonometric functions: $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,$\sin ^{-1}(-x) = -\sin ^{-1}(x)$,and $	an ^{-1}(-x) = -	an ^{-1}(x)$. Given expression: $E = \cos ^{-1}\left(-\frac{1}{2}ight) - 2 \sin ^{-1}\left(\frac{1}{2}ight) + 3 \cos ^{-1}\left(-\frac{1}{\sqrt{2}}ight) - 4 	an ^{-1}(-1)$ $E = \left(\pi - \cos ^{-1}\left(\frac{1}{2}ight)ight) - 2\left(\frac{\pi}{6}ight) + 3\left(\pi - \cos ^{-1}\left(\frac{1}{\sqrt{2}}ight)ight) - 4\left(-\frac{\pi}{4}ight)$ $E = \left(\pi - \frac{\pi}{3}ight) - \frac{\pi}{3} + 3\left(\pi - \frac{\pi}{4}ight) + \pi$ $E = \frac{2\pi}{3} - \frac{\pi}{3} + 3\left(\frac{3\pi}{4}ight) + \pi$ $E = \frac{\pi}{3} + \frac{9\pi}{4} + \pi$ $E = \frac{4\pi + 27\pi + 12\pi}{12} = \frac{43\pi}{12}$

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals:

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