Question

${\cot }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=$ _________

• A. $\frac{\pi }{2}-\frac{1}{2}{\cot }^{-1}x$
• B. ${\cot }^{-1}x$
• C. $-\frac{1}{2}{\tan }^{-1}x$
• D. $\frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}x$

Answer 1

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

${\cot }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
Let $x=\tan \theta$
${\cot }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)={\cot }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$={\cot }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)={\cot }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\pi }{2}-\frac{\theta }{2}=\frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}x$