Question

निम्नलिखित फलन को सरलतम रूप में लिखिए:

${\displaystyle {\tan }^{-1} \frac{x}{\sqrt{a^2-x^2}}}$, |x| < a

Answer 1

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

Put ${\displaystyle x=a\sin \theta }$

${\displaystyle \Rightarrow \frac{x}{a}=\sin \theta }$

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

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

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

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

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

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

${\displaystyle =\theta }$

${\displaystyle ={\sin }^{-1} \frac{x}{a}}$