Question

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

• A. $\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$
• B. $\frac{\pi }{4}+{\cos }^{-1}{x}^2$
• C. $\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x$
• D. $\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}{x}^2$

Answer 1

The correct option is A $\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$

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

Take $x^2=\cos \theta$

$\Rightarrow I={\tan }^{-1}[\frac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }}$

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

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

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

$=\frac{\pi }{4}+\frac{\theta }{2}=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$.