Question

The solution of the equation ${\sin }^{-1}\left(\tan \frac{\pi }{4}\right)-{\sin }^{-1}\left(\sqrt{\frac{3}{x}}\right)-\frac{\pi }{6}=0$ is

• A. $x=2$
• B. $x=-4$
• C. $x=4$
• D. $Noneofthese$

Answer 1

The correct option is C $x=4$

${\sin }^{-1}\left(\tan \frac{\pi }{4}\right)-{\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)-\frac{\pi }{6}=0$

$={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)-\frac{\pi }{6}=0$ $(\because \tan \frac{\pi }{4}=1)$

$=\frac{\pi }{2}-{\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)-\frac{\pi }{6}=0(\because {\sin }^{-1}(1)=\frac{\pi }{2})$

$={\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)=\frac{\pi }{3}$

$\sin \left[{\sin }^{-1}\left(\sqrt{\frac{3}{4}}\right)\right]=\sin \left(\frac{\pi }{3}\right)$

$\sqrt{\frac{3}{x}}=\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{4}}$

$\therefore x=4$