Q. 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
Explanation
(b)
Given, $\sin^{- 1}{x - \cos^{- 1}{x = \frac{\pi}{6}}}$
$\Rightarrow \ \ \left( \frac{\pi}{2} - \cos^{- 1}x \right) - \cos^{- 1}{x = \frac{\pi}{6}}$
$\Rightarrow \ \ \cos^{- 1}{x = \frac{\pi}{6}}$
$\Rightarrow \ \ x = \frac{\sqrt{3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.