Question:
Write the function in the simplest form:
$\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$
Solution:
$\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}$
Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$
$\therefore \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)$
$=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$
$=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$
$=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x$
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