If alpha = 3 sin^{-1}left(frac{6}{11}right) a | vedclass

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Question

(D) Given $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$. Since $\sin^{-1}\left(\frac{1}{2}\right) < \sin^{-1}\left(\frac{6}{11}\right) < \sin^{-1}\l

Vedclass · Accepted Answer

$(B), (C), (D)$

Vedclass · Answer

(D) Given $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$. Since $\sin^{-1}\left(\frac{1}{2}\right) Multiplying by $3$,we get $\frac{\pi}{2} Given $\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)$. Since $\cos^{-1}\left(\frac{1}{2}\right)  \frac{\pi}{3}$.Thus,$\pi In this interval,$\cos \beta Now,$\frac{\pi}{2} Adding these,$\frac{3\pi}{2} In the interval $(\frac{3\pi}{2}, 2\pi)$,$\cos(\alpha + \beta) > 0$. In $(2\pi, \frac{5\pi}{2})$,$\cos(\alpha + \beta) > 0$.Therefore,$\cos(\alpha + \beta) > 0$.The correct options are $(B), (C), (D)$.

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