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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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