Question

If $\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\sin }^{-1}x$, then the value of $x$ is

• A. $\frac{\sqrt{2}}{3}$
• B. $\frac{2\sqrt{2}}{3}$
• C. $2\sqrt{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$LHS=\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}\left(\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{3}\right)\right)$
$=\frac{9}{4}\left({\cos }^{-1}\left(\frac{1}{3}\right)\right)\left(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\right)$
$=\frac{9}{4}{\sin }^{-1}\sqrt{1-{\left(\frac{1}{3}\right)}^2}\left(\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}\right)$
$=\frac{9}{4}{\sin }^{-1}\sqrt{1-\frac{1}{9}}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$
$\therefore \frac{9}{4}{\sin }^{-1}x=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$
$\Rightarrow x=\frac{2\sqrt{2}}{3}$