Question

Write the function in the simplest form:
${\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}},\left|x\right|<a$

Answer 1

${\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}$

Put $x=a\sin \theta \Rightarrow \frac{x}{a}=\sin \theta \Rightarrow \theta ={\sin }^{-1}\left(\frac{x}{a}\right)$

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

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

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