If alpha=3 sin ^{-1} frac{6}{11} and beta=3 c | vedclass

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(A) Given $\alpha = 3 \sin^{-1} \left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1} \left(\frac{4}{9}\right)$.Since $\frac{6}{11} > \frac{1}{2}$,and $

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$\cos \beta > 0$

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(A) Given $\alpha = 3 \sin^{-1} \left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1} \left(\frac{4}{9}\right)$.Since $\frac{6}{11} > \frac{1}{2}$,and $\sin^{-1} x$ is an increasing function,we have $\sin^{-1} \left(\frac{6}{11}\right) > \sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$.Thus,$\alpha = 3 \sin^{-1} \left(\frac{6}{11}\right) > 3 \left(\frac{\pi}{6}\right) = \frac{\pi}{2}$.Also,since $\frac{6}{11} Therefore,$\cos \alpha For $\beta$,since $\frac{4}{9}  \cos^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{3}$.Thus,$\beta = 3 \cos^{-1} \left(\frac{4}{9}\right) > 3 \left(\frac{\pi}{3}\right) = \pi$.Since $\frac{4}{9} > 0$,$\beta In the $III^{rd}$ quadrant,$\cos \beta Since $\alpha \in (\frac{\pi}{2}, \pi)$ and $\beta \in (\pi, \frac{3\pi}{2})$,their sum $\alpha + \beta \in (\frac{3\pi}{2}, \frac{5\pi}{2})$.In this range,$\cos(\alpha + \beta)$ can be positive (in the $IV^{th}$ quadrant) or negative (in the $III^{rd}$ quadrant).Comparing the options: $\cos \beta  0$ is false),$\sin \beta  0$ (true),$\cos \alpha The incorrect option is $A$.

If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the incorrect option is

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