Question

The value of ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ is equal to

• A. ${\cot }^{-1}x$
• B. ${\sec }^{-1}x$
• C. ${\tan }^{-1}x$
• D. none of these

Answer 1

The correct option is D none of these

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

Let $x=\tan A$
Then the above expression reduces to
${\tan }^{-1}\left(\frac{\sec A-1}{\tan A}\right)$
$={\tan }^{-1}\left(\frac{1-\cos A}{\sin A}\right)$
$={\tan }^{-1}\left(\tan \right(\frac{A}{2}\left)\right)$ $=\frac{A}{2}$
$=\frac{1}{2}ta{n}^{-1}x$