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. None of these

Answer 1

The correct option is B

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

Explanation for the correct option.

Given that, ${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6}....\left(1\right)$

Since, ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$

so, Putting ${\sin }^{-1}x=\frac{\mathrm{\pi }}{2}-{\cos }^{-1}x$ in equation (1) we have:

$\begin{array}{rcl}\frac{\mathrm{\pi }}{2}-{\cos }^{-1}x-{\cos }^{-1}x& =& \frac{\mathrm{\pi }}{6}\\ 2{\cos }^{-1}x& =& \frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}\\ 2{\cos }^{-1}x& =& \frac{3\mathrm{\pi }-\mathrm{\pi }}{6}\\ 2{\cos }^{-1}x& =& \frac{\mathrm{\pi }}{3}\\ {\cos }^{-1}x& =& \frac{\mathrm{\pi }}{6}\\ x& =& \frac{\sqrt{3}}{2}\end{array}$

Hence, option B is correct.