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