Question

Show that ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$ for $-\frac{1}{\sqrt{2}}\ge x\ge 1$.

Answer 1

$ta{n}^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=ta{n}^{-1}\frac{1-\sqrt{1-x^2}}{x}$

Put x = sin$\theta$ $\theta \in (-\frac{\pi }{4},\frac{\pi }{2}$)

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

Since $\frac{\theta }{2}\in (\frac{-\pi }{8},\frac{\pi }{4}$)

We can write $ta{n}^{-1}tan\frac{\theta }{2}=\frac{\theta }{2}=\frac{si{n}^{-1}x}{2}=\frac{\frac{\pi }{2}-co{s}^{-1}x}{2}=\frac{\pi }{2}-\frac{1}{2}co{s}^{-1}x$

So the rquired identity is proved .