Question

Statement I The simplest form of $\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}},|x| < a$ is $\sin ^{-1} \frac{x}{a}$.
Statement II The simplest form of $\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), a>0, \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$ is $3 \tan ^{-1} x$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (A)

I. Let $x=a \sin \theta$, then $\sin ^{-1}\left(\frac{x}{a}\right)=\theta$
$ \tan ^{-1} \frac{x}{\sqrt{a^2-x^2}}=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right)$
$=\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^2 \theta}}\right)=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)$
$ \left( \because \sin ^2 x+\cos ^2 x=1 \Rightarrow \cos x=\sqrt{1-\sin ^2 x}\right) $
$ =\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1}\left(\frac{x}{a}\right) $
II. Let $x=a \tan \theta$
$ \Rightarrow \frac{x}{a}=\tan \theta \Rightarrow \theta=\tan ^{-1}\left(\frac{x}{a}\right) ..... $(i)
$ \therefore \tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left[\frac{3 a^2(a \tan \theta)-(a \tan \theta)^3}{a^3-3 a(a \tan \theta)^2}\right]$
$ =\tan ^{-1}\left(\frac{a^3\left(3 \tan \theta-\tan ^3 \theta\right)}{a^3\left(1-3 \tan ^2 \theta\right)}\right) $
$=\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)=\tan ^{-1}(\tan 3 \theta)$
$\left(\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$
$ =3 \theta=3 \tan ^{-1}\left(\frac{x}{a}\right) $[from Eq.(i)]