Question

If $\alpha =3{\sin }^{-1}\frac{6}{11}$ and $\beta =3{\cos }^{-1}\frac{4}{9}$ where the inverse trigonometric functions take 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)

The correct options are
B $\cos (\alpha +\beta )>0$
C $\cos \alpha <0$
D $\sin \beta <0$

$\alpha =3{\sin }^{-1}\frac{6}{11}>3{\sin }^{-1}\frac{6}{12}>3{\sin }^{-1}\frac{1}{2}>\frac{\pi }{2}$
$\beta =3{\cos }^{-1}\frac{4}{9}>3{\cos }^{-1}\frac{4}{8}>3{\cos }^{-1}\frac{1}{2}>\pi$
$\therefore \alpha +\beta >\frac{3\pi }{2}$
Hence $\sin \beta <0,\cos \alpha <0,\cos (\alpha +\beta )>0$