Question

If $si{n}^{-1}x-co{s}^{-1}x=\frac{\pi }{6}$, then the value of x is equal to

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

Answer 1

The correct option is D None of these

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

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

Take $\sin$ on both sides.

$x=\sin ({\cos }^{-1}x+\frac{\pi }{6})=\sin \left({\cos }^{-1}x\right)cos\left(\frac{\pi }{6}\right)+sin\left(\frac{\pi }{6}\right)cos\left(co{s}^{-1}x\right)$

let $co{s}^{-1}x=y$ then $\cos y=x,\sin y=sin\left({\cos }^{-1}x\right)=\pm \sqrt{1-x^2}$

$x=\pm \sqrt{1-x^2}\frac{\sqrt{3}}{2}+\frac{x}{2}$

$\frac{x}{2}=\pm \sqrt{1-x^2}\frac{\sqrt{3}}{2}$

$x^2=3(1-x^2)\Rightarrow x^2=\frac{3}{4}\Rightarrow x=\pm \frac{\sqrt{3}}{2}$