Question

Show that ${\tan }^{-1}\sqrt{x}$=$\frac{1}{2}{\cos }^{-1}(\frac{1-x}{1+x}),x\in [0,1]$

Answer 1

Let $ta{n}^{-1}\sqrt{x}=\theta$ .............................. (1)

Then $\sqrt{x}=tan\theta$ [$\theta \in (0,\frac{\pi }{4}$}

We know that $Cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$=$\frac{1-x}{1+x}$

And since $2\theta \in (0,\frac{\pi }{2}$)

We can write $2\theta =co{s}^{-1}\left(\frac{1-x}{1+x}\right)$......................(2)

Using (1) and (2),

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