The simplest form of $\tan ^{-1}\left\{\frac{x}{\sqrt{a^2-x^2}}\right\}$ is, where -a < x < a.
Answer & explanation
Correct answer: option 4
$y=\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$
Let x = a sin θ
so $\frac{x}{a} = sin θ ⇒ \theta=\sin ^{-1}\left(\frac{x}{a}\right)$
$y=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right) \Rightarrow y=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2} \sqrt{1-\sin ^2 \theta}}\right)$
$y=\tan ^{-1}\left(\frac{a}{a} \frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\right)$
as $\sin ^2 \theta \cos ^2 \theta=1$
$\cos ^2 \theta=1-\sin ^2 \theta$
$y=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta)$
$y=\theta=\sin ^{-1}\left(\frac{x}{a}\right)$