The value of sin^{-1}(sin frac{7pi}{6}) + cos | vedclass

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Question

(A) Step $1$: Simplify $\sin^{-1}(\sin \frac{7\pi}{6})$. Since $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, $\sin(\frac{7\pi}{6}) = -\sin(\frac{\pi}{6}) =

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$\frac{2\pi}{3}$

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(A) Step $1$: Simplify $\sin^{-1}(\sin \frac{7\pi}{6})$. Since $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, $\sin(\frac{7\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$. Thus, $\sin^{-1}(\sin \frac{7\pi}{6}) = -\frac{\pi}{6}$.Step $2$: Simplify $\cos^{-1}(\cos \frac{7\pi}{6})$. Since $\frac{7\pi}{6} = 2\pi - \frac{5\pi}{6}$, $\cos(\frac{7\pi}{6}) = \cos(\frac{5\pi}{6})$. Thus, $\cos^{-1}(\cos \frac{7\pi}{6}) = \frac{5\pi}{6}$.Step $3$: Simplify $	an^{-1}(	an \frac{2\pi}{3})$. Since $\frac{2\pi}{3} = \pi - \frac{\pi}{3}$, $	an(\frac{2\pi}{3}) = -	an(\frac{\pi}{3}) = 	an(-\frac{\pi}{3})$. Thus, $	an^{-1}(	an \frac{2\pi}{3}) = -\frac{\pi}{3}$.Step $4$: Sum the values: $-\frac{\pi}{6} + \frac{5\pi}{6} - \frac{\pi}{3} = \frac{4\pi}{6} - \frac{2\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.Wait, re-evaluating: $\cos^{-1}(\cos \frac{7\pi}{6}) = \cos^{-1}(\cos(2\pi - \frac{5\pi}{6})) = \frac{5\pi}{6}$. Sum is $-\frac{\pi}{6} + \frac{5\pi}{6} - \frac{\pi}{3} = \frac{4\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{3}$. Given options do not match. Let's re-check $\cos^{-1}(\cos \frac{7\pi}{6}) = \cos^{-1}(\cos(2\pi - \frac{7\pi}{6})) = \cos^{-1}(\cos \frac{5\pi}{6}) = \frac{5\pi}{6}$. The sum is $\frac{\pi}{3}$. Since $\frac{\pi}{3}$ is not an option, checking the question again: $\sin^{-1}(\sin \frac{7\pi}{6}) = -\frac{\pi}{6}$, $\cos^{-1}(\cos \frac{7\pi}{6}) = \frac{5\pi}{6}$, $	an^{-1}(	an \frac{2\pi}{3}) = -\frac{\pi}{3}$. Sum $= \frac{4\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{3}$. If the question intended $\cos^{-1}(\cos \frac{7\pi}{6})$ as $\frac{7\pi}{6}$ is outside range, the result is $\frac{5\pi}{6}$. The sum is $\frac{\pi}{3}$. None of the options match.

The value of $\sin^{-1}(\sin \frac{7\pi}{6}) + \cos^{-1}(\cos \frac{7\pi}{6}) + \tan^{-1}(\tan \frac{2\pi}{3})$ is equal to...

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