Question

Write the following in the simplest form.

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

Answer 1

Let $x=tan\theta ,then\theta =ta{n}^{-1}x$
$\therefore ta{n}^{-1}\frac{\sqrt{1+x^2}-1}{x}=ta{n}^{-1}\frac{\sqrt{1+ta{n}^2\theta }-1}{tan\theta }=ta{n}^{-1}\frac{\sqrt{se{c}^2\theta }-1}{tan\theta }$

$[\because 1+ta{n}^2\theta =se{c}^2\theta ]$

$=ta{n}^{-1}\left[\frac{\frac{1}{cos\theta }-1}{\frac{sin\theta }{cos\theta }}\right]=ta{n}^{-1}[\frac{1-cos\theta }{cos\theta }\times \frac{cos\theta }{sin\theta }]$

$ta{n}^{-1}\left[\frac{1-cos\theta }{sin\theta }\right]=ta{n}^{-1}\left[\frac{2i{n}^2\frac{\theta }{2}}{2sin\frac{\theta }{2}cos\frac{\theta }{2}}\right]\left[\begin{array}{c}\because 1-cos\theta =2si{n}^2\frac{\theta }{2}\\ andsin\theta =2sin\frac{\theta }{2}cos\frac{\theta }{2}\end{array}\right]$

$=ta{n}^{-1}\left[\frac{2sin\frac{\theta }{2}}{2cos\frac{\theta }{2}}\right]=ta{n}^{-1}\left[tan\frac{\theta }{2}\right]=\frac{\theta }{2}=\frac{ta{n}^{-1}x}{2}$

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