JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Inverse Trigonometric Functions

Since ${\sin }^{-1}(6/11)\in (0,\pi /2)$, we have $\alpha =3{\sin }^{-1}(6/11)\in (0,3\pi /2)$. Since $6/11>1/2$, ${\sin }^{-1}(6/11)>\pi /6$, so $\alpha >\pi /2$, placing $\alpha$ in the second or third quadrant where $\cos \alpha <0$ (D is correct). Similarly, ${\cos }^{-1}(4/9)\in (0,\pi )$, so $\beta =3{\cos }^{-1}(4/9)\in (0,3\pi )$. Since $4/9<1/2$, ${\cos }^{-1}(4/9)>\pi /3$, so $\beta >\pi$, placing $\beta$ in the third or fourth quadrant. For $\beta \in (\pi ,2\pi )$, $\sin \beta <0$ (B is correct). Detailed calculation shows $\cos (\alpha +\beta )>0$ (C is correct), while $\cos \beta$ can be negative (A is incorrect).