Q. If $- \frac{\pi}{2} < \sin^{- 1}{x < \frac{\pi}{2}}$, then $\tan{(\sin^{- 1}x)}$ is equal to
A) $\frac{x}{1 - x^{2}}$
B) $\frac{x}{1 + x^{2}}$
C) $\frac{x}{\sqrt{1 - x^{2}}}$
D) $\frac{1}{\sqrt{1 - x^{2}}}$
Explanation
(c)
$\tan{\left( \sin^{- 1}x \right) = \tan{\left( \tan^{- 1}\frac{x}{\sqrt{1 - x^{2}}} \right),\ x \in ( - 1,\ 1)}}$
$= \frac{x}{\sqrt{1 - x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.