Question

If ${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6},$ then $x$ is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $-\frac{1}{2}$
• D. none of these.

Answer 1

Answer: (B)

$si{n}^{-1}x-co{s}^{-1}x=\frac{\pi }{6}$
$\Rightarrow \frac{\pi }{2}-co{s}^{-1}x-co{s}^{-1}x=\frac{\pi }{6}$
$\Rightarrow 2co{s}^{-1}x=\frac{\pi }{2}-\frac{\pi }{6}$
$=\frac{2\pi }{6}-\frac{\pi }{3}$
$\Rightarrow co{s}^{-1}=\frac{\pi }{6}$
$\therefore x=cos\frac{\pi }{6}$
$=\frac{\sqrt{3}}{2}$