Question

Evaluate ${\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)$, where $x\ne 0$

• A. $\frac{{\tan }^{-1}ax}{2}$
• B. $\frac{{\cot }^{-1}ax}{2}$
• C. $\frac{{\cot }^{-1}ax}{4}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{{\tan }^{-1}ax}{2}$

Put $ax=\tan \theta$
$\therefore {\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}}\right)$
$={\tan }^{-1}\tan \frac{\theta }{2}=\frac{\theta }{2}=\frac{{\tan }^{-1}\left(ax\right)}{2}$