Question

Write the simplest form of :

$ta{n}^{-1}\frac{\sqrt{1+x^2}-1}{x},x\ne 0$

Answer 1

Put $x=\tan \theta$

$\theta =ta{n}^{-1}x$

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

$=ta{n}^{-1}\frac{\sqrt{1+ta{n}^2\theta }-1}{tan\theta }$

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

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

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