Question

If $\alpha ={\sin }^{-1}\frac{\sqrt{3}}{2}+{\sin }^{-1}\frac{1}{3}$ and $\beta ={\cos }^{-1}\frac{\sqrt{3}}{2}+{\cos }^{-1}\frac{1}{3}$, then

• A. $\alpha >\beta$
• B. $\alpha =\beta$
• C. $\alpha <\beta$
• D. $\alpha +\beta =2\pi$

Answer 1

Answer: (C)

$\alpha <\beta$

From the given equation, it can be seen that $\alpha +\beta =\pi$ .........(1)

Since, ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\forall x$

Also, $\alpha =\frac{\pi }{3}+{\sin }^{-1}\frac{1}{3}<\frac{\pi }{3}+{\sin }^{-1}\frac{1}{2}$

as $\sin \theta$ is increasing in $[0,\frac{\pi }{2}]$

$\therefore \alpha <\frac{\pi }{3}+\frac{\pi }{6}=\frac{\pi }{2}$

$\Rightarrow \beta >\frac{\pi }{2}>\alpha \Rightarrow \alpha <\beta$ [from (1)}