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 ${\tan }^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right),a>0,\frac{-a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}$ 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)
$\therefore {\tan }^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right)={\tan }^{-1}\left[\frac{3a^2(a\tan \theta )-(a\tan \theta {)}^3}{a^3-3a(a\tan \theta {)}^2}\right]$
$={\tan }^{-1}\left(\frac{a^3\left(3\tan \theta -{\tan }^3\theta \right)}{a^3\left(1-3{\tan }^2\theta \right)}\right)$
$={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)={\tan }^{-1}(\tan 3\theta )$
$\left(\because \tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$
$=3\theta =3{\tan }^{-1}\left(\frac{x}{a}\right)$[from Eq.(i)]