STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

$\therefore \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}$

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

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

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