Q. If $x$ takes negative permissible value, then $\sin^{- 1}x$ is equal to
A) $- \cos^{- 1}\sqrt{1 - x^{2}}$
B) $\cos^{- 1}\sqrt{x^{2} - 1}$
C) $\pi - \cos^{- 1}\sqrt{1 - x^{2}}$
D) $\cos^{- 1}\sqrt{1 - x^{2}}$
Explanation
(a)
Let $x = - y,\ y > 0$
$\therefore\sin^{- 1}{x = \sin^{- 1}( - y)}$
$= {- sin}^{- 1}y$
$= - \cos^{- 1}\sqrt{1 - y^{2}}$
$= - \cos^{- 1}\sqrt{1 - x^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.