Question

Write the simplest form of ${\tan }^{-1}\left[\frac{x}{\sqrt{a^2-x^2}}\right],|x|<a$

Answer 1

${\tan }^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$

Put $x=a\sin \theta$

$={\tan }^{-1}\left(\frac{a\sin \theta }{\sqrt{a^2-a^2{\sin }^2\theta }}\right)$

$={\tan }^{-1}\left(\frac{a\sin \theta }{\sqrt{a^2(1-{\sin }^2\theta )}}\right)$

$={\tan }^{-1}\left(\frac{a\sin \theta }{a\sqrt{(1-{\sin }^2\theta )}}\right)$

$={\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{\left({\cos }^2\theta \right)}}\right)$

$={\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right)$

$={\tan }^{-1}\left(\tan \theta \right)$

$=\theta$

$={\sin }^{-1}\frac{x}{a}$ since $x=a\sin \theta \Rightarrow \theta ={\sin }^{-1}\frac{x}{a}$

Hence,${\tan }^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)={\sin }^{-1}\frac{x}{a}$