If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$, where the inverse…
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
$\cos \beta>0$
$\sin \beta < 0$
$\cos (\alpha+\beta)>0$
$\cos \alpha < 0$
Solution
$\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$
Since $\frac{6}{11}>\frac{6}{12}$
Taking $\sin ^{-1}$ on both sides, we get
$\sin ^{-1}\left(\frac{6}{11}\right)>\sin ^{-1}\left(\frac{6}{12}\right)$
$\ldots\left[\because \sin ^{-1} x\right.$ is an increasing function $]$
$\Rightarrow 3 \sin ^{-1}\left(\frac{6}{11}\right)>3 \sin ^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \alpha>3\left(\frac{\pi}{6}\right)$
$\Rightarrow \alpha>\frac{\pi}{2}$
Now, $\frac{4}{9} < \frac{4}{8}$
Taking $\cos ^{-1}$ on both sides, we get $\cos ^{-1}\left(\frac{4}{9}\right)>\cos ^{-1}\left(\frac{4}{8}\right)$
$\ldots\left[\because \cos ^{-1} x\right.$ is a decreasing function $]$
$\Rightarrow 3 \cos ^{-1}\left(\frac{4}{9}\right)>3 \cos ^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \beta>3\left(\frac{\pi}{3}\right)$
$\Rightarrow \beta>\pi$
From (i) and (ii), we have $\alpha$ lies in II $^{\text {nd }}$ quadrant and $\beta$ lies in III $^{\text {rd }}$ quadrant.
$\therefore \quad \cos \alpha < 0, \cos \beta < 0 \text { and } \sin \beta < 0$
Also, $\alpha+\beta>\frac{\pi}{2}+\pi$
....[From (i) and (ii)]
$\therefore \quad \alpha+\beta>\frac{3 \pi}{2}$
Thus, $\alpha+\beta$ lies in IV $^{\text {th }}$ quadrant.
So, $\cos (\alpha+\beta)>0$
[Note: Options (B), (C) and (D) are correct.]