Question

If $\alpha =3{\sin }^{-1}(\frac{6}{11})$ and $\beta =3{\cos }^{-1}(\frac{4}{9}),$ where the inverse trignometric functions takes only the principal values, then the correct option(s) is(are)

• A. $\cos \beta >0$
• B. $\sin \beta <0$
• C. $\cos (\alpha +\beta )>0$
• D. $\cos \alpha <0$

Answer 1

Answer: (B), (C), (D)

$\cos \alpha <0$

$\alpha =3{\sin }^{-1}(\frac{6}{11})$
$\frac{\alpha }{3}={\tan }^{-1}(\frac{6}{\sqrt{85}})$
As, $\frac{1}{\sqrt{3}}<\frac{6}{\sqrt{85}}<1\frac{\pi }{6}<\frac{\alpha }{3}<\frac{\pi }{4}\frac{\pi }{2}<\alpha <\frac{3\pi }{4}$
$\Rightarrow \cos \alpha <0$
Now,
$\beta =3{\cos }^{-1}(\frac{4}{9})$
$\Rightarrow \frac{\beta }{3}={\tan }^{-1}(\frac{\sqrt{65}}{4})$
As we know $(\frac{\sqrt{65}}{4})>\sqrt{3}$
so, $\frac{\pi }{3}<\frac{\beta }{3}<\frac{\pi }{2}\pi <\beta <\frac{3\pi }{2}\Rightarrow \sin \beta <0$
Also, $(\alpha +\beta )$ will lies in the fourth quadrant
$\Rightarrow \cos (\alpha +\beta )>0$