Question

If ${\sin }^{-1}\left(\tan \frac{\pi }{4}\right)-{\sin }^{-1}\sqrt{\frac{3}{x}}=\frac{\pi }{6}$, then $x$ is a root of the equation:

• A. $x^2-x-6=0$
• B. $x^2+x-6=0$
• C. $x^2-x-12=0$
• D. $x^2+x-12=0$

Answer 1

Answer: (C)

$x^2-x-12=0$

${\sin }^{-1}\left(1\right)-\frac{\pi }{6}={\sin }^{-1}\sqrt{\frac{3}{x}}$
$\frac{\pi }{2}-\frac{\pi }{6}={\sin }^{-1}\sqrt{\frac{3}{x}}$
$\sqrt{\frac{3}{x}}=\frac{\sqrt{3}}{2}\Rightarrow x=4$
Hence, option 'C' is correct.