Question

Write the following in simplest form : ${\tan }^{-1}\left(\frac{\sqrt{(1+x^2)}-1}{x}\right)$

Answer 1

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

Put $x=\tan \theta$

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

Therefore, the function becomes

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

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

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

$=ta{n}^{-1}\frac{2si{n}^2\frac{\theta }{2}}{2sin\frac{\theta }{2}cos\frac{\theta }{2}}$

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

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

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