Question

Prove that: $ta{n}^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]=\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x,\frac{-1}{\sqrt{2}}\le x\le 1$

Answer 1

Let $x=\cos \theta ,\text{so that}{\cos }^{-1}x=\theta$

Now, ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

$={\tan }^{-1}\left(\frac{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}\right)$

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

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

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

$={\tan }^{-1}(\tan \frac{\pi }{4}-\frac{\theta }{2})=\frac{\pi }{4}-\frac{\theta }{2}$

$(\because \frac{-1}{\sqrt{2}}\le x\le 1\Rightarrow \frac{-1}{\sqrt{2}}\le \cos \theta \le 1\Rightarrow cos\frac{3\pi }{4}\le \cos \theta \le \cos 0\Rightarrow \frac{3\pi }{4}\le \theta \le 0\Rightarrow 0\le \theta \le \frac{3\pi }{4}\Rightarrow \frac{-3\pi }{8}\le \frac{-\theta }{2}\le 0\Rightarrow \frac{\pi }{4}+\frac{-3\pi }{8}\le \frac{\pi }{4}+\frac{-\theta }{2}\le \frac{\pi }{4}\Rightarrow -\frac{\pi }{4}\le \frac{\pi }{4}-\frac{\theta }{2}\le \frac{\pi }{4})$

$=\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x$