2021/004) Given $(\sqrt{x+\sqrt{x^2+1}})(\sqrt{y+\sqrt{y^2+1}}) =1 $ find $(x+y)^2$

Let $x=\tan\theta$ $\theta$ in the range $0 ..\frac{\pi}{2}$ so we have $sec\theta$ positive

Then we have $\sqrt{x+\sqrt{x^2+1}} = \sqrt{\tan\theta + \sec\theta} $

From the given condition we have

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\sqrt{x+\sqrt{x^2+1}}}$

Or  

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\tan \theta + \sec\theta}$

$=\frac{\sec^\theta - tan^2\theta }{\tan \theta + \sec\theta}$

or $y+\sqrt{y^2+1} =\sec\theta - \tan \theta\cdots(1)$

if we take $y=\tan\alpha$ $\alpha$ in the range $-\frac{\pi}{2} ..\frac{\pi}{2}$ so we have $sec,\theta$ positive

then so $y + \sqrt{y^2+1}= \sec\alpha \pm \tan \alpha\cdots(2)$

comparing (1) and (2) we have $y =-\tan \theta$

so $x+y=0$ or $(x+y)^2 = 0$