Question

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

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $-\frac{1}{2}$
• D. $-\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

We have,

${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6}$

We know that

${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$

Thus,

${\sin }^{-1}x-(\frac{\pi }{2}-{\sin }^{-1}x)=\frac{\pi }{6}$

${\sin }^{-1}x-\frac{\pi }{2}+{\sin }^{-1}x=\frac{\pi }{6}$

$2{\sin }^{-1}x=\frac{\pi }{6}+\frac{\pi }{2}$

$2{\sin }^{-1}x=\frac{2\pi }{3}$

${\sin }^{-1}x=\frac{\pi }{3}$

$x=\sin \frac{\pi }{3}$

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

Hence, this is the answer.