Sunday, January 24, 2021

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 $(x+y)^2 = 0$


 

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