Question

Solve ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x},x\ne 0$.

• A. $\frac{{\cot }^{-1}x}{2}$
• B. $\frac{{\tan }^{-1}x}{2}$
• C. $\frac{{\sin }^{-1}x}{2}$
• D. None of these

Answer 1

The correct option is B $\frac{{\tan }^{-1}x}{2}$

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

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

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

$={\tan }^{-1}\frac{\sec \theta -1}{\tan \theta }$

$={\tan }^{-1}\frac{1-\cos \theta }{\sin \theta }$

$={\tan }^{-1}\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$

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

$=\frac{\theta }{2}$

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