Question

Prove that $\frac{1}{2}{\tan }^{-1}x={\cos }^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right]$

Answer 1

${\cos }^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right]$

put $x=\tan \theta$

$={\cos }^{-1}\left[\sqrt{\frac{1+\sqrt{1+{\tan }^2\theta }}{2\sqrt{1+{\tan }^2\theta }}}\right]$

$={\cos }^{-1}\left[\sqrt{\frac{1+\sqrt{{\sec }^2\theta }}{2\sqrt{{\sec }^2\theta }}}\right]$

$={\cos }^{-1}\left[\sqrt{\frac{1+\sec \theta }{2\sec \theta }}\right]$

$={\cos }^{-1}\left[\sqrt{\frac{1+\cos \theta }{2}}\right]$

$={\cos }^{-1}\left[\sqrt{{\cos }^2\frac{\theta }{2}}\right]$

$={\cos }^{-1}\left[\cos \frac{\theta }{2}\right]$

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

$x=\tan \theta$

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

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

$\therefore \frac{1}{2}{\tan }^{-1}x={\cos }^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right]$